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Problem 4.9.28. Let \( {ABC} \) be a triangle with incenter \( I \) . The points \( F, D \) and \( E \) are chosen on the perpendicular lines from \( I \) to the sides \( {AB} \) , \( {BC} \) and \( {CA} \), such that their distances to \( I \) are equal. Prove that the lines \( {AD},{BE} \) and \( {CF} \) are concurre... | Solution. Let the incircle of \( \bigtriangleup {ABC} \) touch the sides \( {BC},{CA} \) and \( {AB} \) at the points \( M, N \) and \( P \), respectively. Then it is easy to see that \( \bigtriangleup {CMD} \cong \bigtriangleup {CNE},\bigtriangleup {ANE} \cong \bigtriangleup {APF} \) and \( \bigtriangleup {BPF} \cong ... | Yes |
Problem 4.9.29. Let \( {ABC} \) be a triangle. Let \( X \) be an arbitrary point, not lying on any of the sides of the triangle. Denote \( D = {AX} \cap {BC} \) . \( E = {BX} \cap {AC} \) and \( F = {CX} \cap {AB} \) . The circles \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) are external to \( \bigtriangleup {ABC} \) . They... | Solution. By Problem 6.1.1 we have that \( {MD} \) passes through the midpoint of the arc \( \overset{⏜}{BAC} \) and hence it bisects \( \angle {BMC} \) .\n\nAnalogously, \( {NE} \) bisects \( \angle {ANC} \) and \( {PF} \) bisects \( \angle {APB} \) .\n\nWe have\n\n be a triangle with circumcircle \( k \) . Let \( X \) be an arbitrary point, not lying on any of the sides of the triangle. Denote \( {XA} \cap k = \left\{ {X,{A}_{1}}\right\} ,{XB} \cap k = \left\{ {X,{B}_{1}}\right\} \) and \( {XC} \cap k = \left\{ {X,{C}_{1}}\right\} \) . The points \... | Solution. If \( X \) lies\non \( k \), the statement is clearly true. Let \( X \notin \) \( k \) . Denote the circumcircles of \( \bigtriangleup X{A}_{1}D,\bigtriangleup X{B}_{1}E \) and \( \bigtriangleup X{C}_{1}F \) by \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) , respectively.\n\nIt suffices to show that two of the circ... | Yes |
Problem 4.10.1. Let \( {ABC} \) be a triangle with altitude \( {CH}\left( {H \in {AB}}\right) \) . Let \( {CL}\left( {L \in {AB}}\right) \) be the angle bisector of \( \angle {ACH} \) . Let the line through \( B \), that is parallel to \( {CL} \), intersect \( {CH} \) at the point \( N \) . Let \( M \) be the midpoint ... | Solution. If \( \angle {BAC} = \alpha \), then \( \angle {ABC} = {90}^{ \circ } - \alpha ,\angle {LCH} = {45}^{ \circ } - \frac{\alpha }{2} \) , \( \angle {CLH} = {45}^{ \circ } + \frac{\alpha }{2} \) and \( \angle {LCB} = {45}^{ \circ } + \frac{\alpha }{2} \) . Therefore, \( \bigtriangleup {LBC} \) is isosceles. Consi... | Yes |
Problem 4.10.2. Let \( {ABC} \) be a triangle with \( \angle {ACB} = {90}^{ \circ } \), and let \( {CD}\left( {D \in {AB}}\right) \) be an altitude. The incenters of \( \bigtriangleup {ADC} \) and \( \bigtriangleup {BDC} \) are \( E \) and \( F \), respectively. Prove that the quadrilateral \( {ABFE} \) is cyclic. | Solution. Denote \( \angle {ABF} = \alpha \) . Now, \( \angle {ABC} = {2\alpha } \) and \( \angle {ACD} = \angle {ABC} = {2\alpha } \). We have that \( \angle {AEC} = {135}^{ \circ } \) . The law of sines, applied to \( \bigtriangleup {AEC} \) yields \[ \frac{AC}{\sin {45}^{ \circ }} = \frac{CE}{\sin \left( {{45}^{ \ci... | Yes |
Problem 4.10.3. Let \( {ABC} \) be a triangle with \( \angle {ACB} = {90}^{ \circ } \), and let \( {CH}\left( {H \in {AB}}\right) \) be an altitude. The incenters of \( \bigtriangleup {AHC} \) and \( \bigtriangleup {BHC} \) are \( {I}_{1} \) and \( {I}_{2} \), respectively. The points \( X \) and \( Y \) lie on the sid... | Solution. Denote \( X{I}_{1} \cap \)\n\n\n\n\( {AB} = {T}_{1} \) . Thus, \( {T}_{1}X\parallel \) \( {BC} \) . The intercept theorem, applied to \( \bigtriangleup {ABC} \) and the points \( {T}_{1} \) and \( X \), yie... | Yes |
Problem 4.10.4. Let \( {ABC} \) be a triangle with \( \angle {ACB} = {90}^{ \circ } \) and angle bisectors \( A{A}_{1}\left( {{A}_{1} \in {BC}}\right) \) and \( B{B}_{1}\left( {{B}_{1} \in {AC}}\right) \), which intersect at the point \( I \) . Prove that the line, defined by \( I \) and the midpoint of the segment \( ... | Solution. Let \( H \) be the foot of the perpendicular from \( I \) to \( {AB} \) and\n\nThe law of sines, applied to \( \bigtriangleup {AIB} \), yields \( \frac{AI}{BI} = \frac{\sin \left( {{45}^{ \circ } - \frac{\alpha }{2}}\right) }{\sin \frac{\alpha }{2}} \) .\n\nThe same theorem, applied to \( \bigtriangleup A{B}_... | No |
Problem 4.10.5. Let \( {ABC} \) be a triangle with \( \angle {ACB} = {90}^{ \circ } \) and incircle \( \omega \), which touches \( {AB},{BC} \) and \( {CA} \) at the points \( N, M \) and \( P \) , respectively. A point \( K \) is chosen on the line \( {AC} \), such that \( {CK} = {BM} \) and \( C \) lies between \( K ... | Solution. Let \( I \) be the center of \( \omega \) . It is easy to see that the quadrilateral \( {IMCP} \) is a square.\n\nSince \( {BM} = {BN} = s - \) \( b \), then \( {CK} = s - b \) and thus \( {AK} = b + s - b = s.\) \n\nFurthermore, \( {CM} = \n\ns - c,{MB} = s - b = {BN} \n\nand \( {AN} = s - a \) . \n\nWe will... | Yes |
Problem 4.10.6. Let \( {ABC} \) be a triangle with \( \angle {BAC} = {90}^{ \circ } \) . Let \( M \) and \( N \) be the midpoints of \( {AC} \) and \( {BC} \), respectively. The point \( D \in {BC} \) is arbitrary. Let \( k \) be the circumcircle of \( \bigtriangleup {ABD} \) . The tangent lines to \( k \) at the point... | Solution. Let \( k \) intersect \( {AC} \) again at the point \( E \), and let \( {DE} \cap {AB} = \) \( P \) . We have that the polar lines of the points \( C, F \) and \( P \) pass through the intersection point of the diagonals of the quadrilateral \( {ABDE} \) . Therefore, the points \( C, F \) and \( P \) are coll... | Yes |
Problem 4.10.7. Let \( {ABC} \) be a triangle with \( \angle {ACB} = {90}^{ \circ } \) and an incircle \( \omega \) that touches \( {AB},{BC} \) and \( {CA} \) at the points \( M, N \) and \( P \) , respectively. Let \( K \) be the reflection of \( P \) with respect to the midpoint of | Solution. Considering \( \bigtriangleup {AMP} \), we have that \( {AP} = {AM} = s - a \) , \( \angle {PAM} = \alpha \) and \( \angle {APM} = {90}^{ \circ } - \frac{\alpha }{2} \) . Let \( T = {PM} \cap {KN} \) . Then \( \angle {KPT} = \) \( {90}^{ \circ } - \frac{\alpha }{2} \)\n\nDenote \( \angle {NKC} = x \) . We hav... | Yes |
Problem 4.10.8. Let \( {ABC} \) be a triangle with \( \angle {ACB} = {90}^{ \circ } \) . Let \( {CD} \) be its altitude. Let \( {I}_{1} \) and \( {I}_{2} \) be the incenters of \( \bigtriangleup {ADC} \) and \( \bigtriangleup {BDC} \) , respectively. Let \( P \) and \( Q \) be the points of tangency of the sides \( {AC... | Solution. Let \( I \) be the incenter of \( \bigtriangleup {ABC} \) . Let the projections of \( {I}_{1} \) and \( {I}_{2} \) onto \( {AB} \) be \( E \) and \( F \), respectively.\n\nWe have that \( \bigtriangleup {BDC} \sim \bigtriangleup {BCA} \) with a ratio of \( \cos \beta \) . The points \( F \) and \( Q \) are co... | Yes |
Problem 4.11.1. Let \( {ABC} \) be a triangle with \( \angle {ACB} = {60}^{ \circ } \) . The angle bisectors \( A{A}_{1}\left( {{A}_{1} \in {BC}}\right) \) and \( B{B}_{1}\left( {{B}_{1} \in {AC}}\right) \) intersect at the point \( I \) . Prove that \( {A}_{1}I = {B}_{1}I \) . | Solution. It is clear that\n\n\n\n\( \angle {A}_{1}I{B}_{1} = {120}^{ \circ } \) .\n\nThen the quadrilateral \( {A}_{1}C{B}_{1}I \) is cyclic. Since \( {CI} \) bisects \( \angle {A}_{1}C{B}_{1} \), it follows that \(... | Yes |
Problem 4.11.2. Let \( {ABC} \) be a triangle with \( \angle {ACB} = {60}^{ \circ } \) . The angle bisectors \( A{A}_{1}\left( {{A}_{1} \in {BC}}\right) \) and \( B{B}_{1}\left( {{B}_{1} \in {AC}}\right) \) intersect at the point \( I \) . Prove that the reflection of \( C \) with respect to the line \( {A}_{1}{B}_{1} ... | Solution. Let \( H \in {A}_{1}{B}_{1} \) be a point such that \( {CH} \bot {A}_{1}{B}_{1} \) and \( {CH} \cap \) \( {AB} = K \) . It suffices to prove that \( {CH} = {HK} \) .\n\nWe have \( \angle {B}_{1}{CH} = {90}^{ \circ } - \angle {A}_{1}{B}_{1}C \) . As in Problem 4.11.1, we see that the quadrilateral \( C{B}_{1}I... | Yes |
Problem 4.11.3. Let \( {ABC} \) be a triangle with \( \angle {ACB} = {60}^{ \circ } \) . The angle bisectors \( A{A}_{1}\left( {{A}_{1} \in {BC}}\right) \) and \( B{B}_{1}\left( {{B}_{1} \in {AC}}\right) \) intersect at the point \( I \) . Prove that the second intersection point of the circumcircles of \( \bigtriangle... | Solution. Let \( K \) be the reflection of \( C \) with respect to \( {A}_{1}{B}_{1} \) . Problem 4.11.2 implies that \( K \) lies on \( {AB} \) . In the proof of 4.11.2 we showed that \( K \) lies on the circumcircles of \( \bigtriangleup A{A}_{1}C \) and \( \bigtriangleup B{B}_{1}C \), as desired. | Yes |
Problem 4.11.4. Let \( {ABC} \) be a triangle with circumcenter \( O \) and \( \angle {ACB} = {60}^{ \circ } \) . Let \( A{A}_{1}\left( {{A}_{1} \in {BC}}\right) \) and \( B{B}_{1}\left( {{B}_{1} \in {AC}}\right) \) be altitudes in the triangle, intersecting at the point \( H \) . The line \( {OH} \) intersects the lin... | Solution. We have that \( \angle {AOB} = \angle {AHB} = {120}^{ \circ } \), so the quadrilateral \( {ABHO} \) is cyclic. Also, \( \angle {BAO} = {30}^{ \circ } \), thus \( \angle {AHO} = {30}^{ \circ } \) .\n\nNote that \( \angle {AH}{B}_{1} = {60}^{ \circ } \) . Therefore, \( \angle {B}_{1}{HP} = {30}^{ \circ } \), wh... | Yes |
Problem 4.11.5. Let \( {ABC} \) be a triangle with \( \angle {ACB} = {120}^{ \circ } \). The lines \( A{A}_{1}\left( {{A}_{1} \in {BC}}\right), B{B}_{1}\left( {{B}_{1} \in {AC}}\right) \) and \( C{C}_{1}\left( {{C}_{1} \in {AB}}\right) \) are its angle bisectors. Prove that \( \angle {A}_{1}{C}_{1}{B}_{1} = {90}^{ \cir... | Solution. It is clear that \( C{A}_{1} \) is an external angle bisector of \( \bigtriangleup {AC}{C}_{1} \). Also, the point \( {A}_{1} \) lies on the bisector of \( \angle {BAC} \). Therefore, \( {A}_{1} \) is the \( A \) -excenter of \( \bigtriangleup {AC}{C}_{1} \). Thus, \( {C}_{1}{A}_{1} \) is the angle bisector o... | Yes |
Problem 4.11.6. Let \( {ABC} \) be a triangle with \( \angle {ACB} = {120}^{ \circ } \) . The lines \( A{A}_{1}\left( {{A}_{1} \in {BC}}\right), B{B}_{1}\left( {{B}_{1} \in {AC}}\right) \) and \( C{C}_{1}\left( {{C}_{1} \in {AB}}\right) \) are its angle bisectors. Prove that \( \angle A{A}_{1}{C}_{1} = B{B}_{1}{C}_{1} ... | Solution. Let \( I \) be the incenter of \( \bigtriangleup {ABC} \) . As in the proof of 4.11.5, we can show that \( {C}_{1}{A}_{1} \) bisects \( \angle B{C}_{1}C \) . Let \( {C}_{1}{A}_{1} \cap {BI} = {I}_{1} \) .\n\n be a circle with center \( C \) . The points \( A \) and \( B \) on \( k \) are such that \( \angle {ACB} = {60}^{ \circ } \) . For an arbitrary point \( D \in \widehat{AB} \), the points \( M, N, P \) and \( Q \) are the midpoints of the segments \( {AD},{DB},{BC} \) and \( {CA} \) . Prove ... | Solution. The given conditions imply that \( \bigtriangleup {ABC} \) is equilateral. Note that \( {PQ} \) and \( {MN} \) are midsegments in \( \bigtriangleup {ABC} \) and \( \bigtriangleup {ABD} \), respectively.\n\nThus, \( {PQ}\parallel {AB}\parallel {MN} \) and \( {PQ} = {MN} = \frac{AB}{2} \) . On the other hand, \... | Yes |
Problem 4.11.8. Let \( {ABC} \) be a triangle with \( \angle {ACB} = {30}^{ \circ } \) . Let \( A{A}_{1}\left( {{A}_{1} \in {BC}}\right) \) and \( B{B}_{1}\left( {{B}_{1} \in {AC}}\right) \) be altitudes in the triangle, and let \( M \) and \( N \) be the midpoints of \( {AC} \) and \( {BC} \), respectively. Prove that... | Solution. Since \( {A}_{1}M \) and \( {B}_{1}N \) are medians to the hypothenuses in \( \bigtriangleup A{A}_{1}C \) and \( \bigtriangleup B{B}_{1}C \), respectively, we have \( \angle N{B}_{1}C = \) \( \angle {ACB} = \angle M{A}_{1}C = {30}^{ \circ } \) .\n\nHence, \( \angle {B}_{1}N{A}_{1} = {60}^{ \circ } \) . Now, i... | Yes |
Problem 4.11.9. Let \( {ABC} \) be a triangle with circumcenter \( O \) and \( \angle {ACB} = {45}^{ \circ } \) . Let \( A{A}_{1}\left( {{A}_{1} \in {BC}}\right) \) and \( B{B}_{1}\left( {{B}_{1} \in {AC}}\right) \) be altitudes in the triangle, and let \( H \) be its orthocenter. Prove that the midpoint of \( {OH} \) ... | Solution. We have that \( \angle {A}_{1}{AC} = {45}^{ \circ } \) and \( \angle {B}_{1}{BC} = \) \( {45}^{ \circ } \) . Thus, \( A{A}_{1} = C{A}_{1} \) and \( {AO} = \) \( {CO} \) . Hence, the line \( {A}_{1}O \) is the perpendicular bisector of \( {AC} \) . In particular, \( {A}_{1}O\parallel H{B}_{1} \) . Analogously,... | Yes |
Problem 4.12.1. Let \( {ABC} \) be a triangle with an altitude \( C{C}_{1}\left( {{C}_{1} \in }\right. \) \( {AB}) \) . An arbitrary point \( P \) is chosen on the side \( {BC} \) . The lines \( {AP} \) and \( C{C}_{1} \) intersect at the point \( E \), and the line \( {BE} \) intersects \( {AC} \) at the point \( Q \)... | Solution. Consider the line \( l \) through \( C \), parallel to \( {AB} \) . Let the lines \( {C}_{1}Q \) and \( {C}_{1}P \) intersect \( l \) at the points \( X \) and \( Y \), respectively.\n\nWe have \( \bigtriangleup A{C}_{1}Q \sim \bigtriangleup {CXQ} \) and \( \bigtriangleup B{C}_{1}P \sim \bigtriangleup {CYP} \... | Yes |
Let \( {ABCD} \) be a quadrilateral. The rays \( {AB} \) and \( {DC} \) intersect at the point \( E \) and the rays \( {DA} \) and \( {CB} \) intersect at the point \( F \) . Let \( {AC} \cap \bar{BD} = P \) and let \( H \) be the foot of the perpendicular from \( P \) to \( {EF} \) . Prove that \( \angle {AHB} = \angl... | Let \( {FP} \cap {DC} = M \) . We apply the law of sines to \( \bigtriangleup {CEH} \) , \( \bigtriangleup {CPH},\bigtriangleup {HAP},\bigtriangleup {HAF} \) and \( \bigtriangleup {EDF} \), as well as Menelaus’ Theorem and Ceva's Theorem, and we get\n\n and \( {BDP} \) have equal angles at the vertex \( B \) and they have the same orientation. Let \( {DP} \cap {AC} = Q \) and \( {AP} \cap {DC} = M \) . Prove that \( \angle {PBM} = \angle {CBQ} \) . | Solution. Let \( \angle {DBP} = \angle {CBA} = \alpha ,\angle {PBM} = {\beta }_{1} \) and \( \angle {QBC} = {\beta }_{2} \) . It suffices to prove that \( \cot {\beta }_{1} = \cot {\beta }_{2} \) .\n\nWe have\n\n\[ \frac{\sin \angle {DBM}}{\sin \angle {PBM}} = \frac{DM}{PM} \cdot \frac{\sin \angle {BDM}}{\sin \angle {B... | Yes |
Let \( D \) and \( E \) be arbitrary points on the sides \( {BC} \) and \( {AC} \) of the triangle \( {ABC} \), respectively. The lines \( {AD} \) and \( {BE} \) intersect at the point \( S \) . Let \( F \in {AB} \) be an arbitrary point, and let \( l \) be a line through \( C \), parallel to \( {AB} \) . The lines \( ... | Note that \( \bigtriangleup {PEC} \sim \bigtriangleup {FEA} \) and \( \bigtriangleup {CDQ} \sim \bigtriangleup {BDF} \) . Thus,\n\n\( \frac{CP}{AF} = \frac{CE}{EA} \) and \( \frac{CQ}{FB} = \frac{CD}{DB}. \)\n\nHence,\n\n ,{BF}\left( {F \in {AC}}\right) \) and \( {CD}\left( {D \in {AB}}\right) \) are the angle bisectors of the triangle \( {ABC} \) . Let \( M = \) \( {AE} \cap {DF} \) and \( N = {BF} \cap {DE} \) . Prove that \( \angle {ACM} = \angle {BCN} \) . | Solution. We assume that the lines \( {EF} \) and \( {AB} \) are not parallel, because otherwise the desired equality easily follows.\n\nLet \( X \) be their intersection point. Problem 4.3.1 yields that \( {CX} \) is the external angle bisector of \( \angle {ACB} \) . Thus, \( \angle {DCX} = {90}^{ \circ } \) .\n\nLet... | Yes |
Problem 4.12.6. The point \( D \) is the midpoint of the side \( {AB} \) of the triangle \( {ABC} \) . The triangles \( {APC} \) and \( {BQC} \) are constructed externally of the triangle, such that \( \angle {APC} = \angle {BQC} = {90}^{ \circ } \) and \( \angle {PAC} = \angle {QBC} \) . Prove that \( {PD} = {QD} \) . | Solution. Let \( N \) and \( M \) be the midpoints of the sides \( {AC} \) and \( {BC} \), respectively. Since \( {DN} \) and \( {DM} \) are midsegments in \( \bigtriangleup {ABC} \), we have \( \angle {AND} = \angle {BMD} = \) \( \gamma \) . Also, the triangles \( \bigtriangleup {PNA} \) and \( \bigtriangleup {QMB} \)... | Yes |
Problem 4.12.7. Let \( {ABC} \) be a triangle. The points \( M \) and \( N \) are external for \( \bigtriangleup {ABC} \) and such that the lines \( {CN} \) and \( {CM} \) are isogonally conjugate with respect to \( \angle {ACB} \), and \( \angle {NAC} = \angle {MBC} = {90}^{ \circ } \) . Let \( {CH}\left( {H \in {AB}}... | Solution. Set \( \angle {NCA} = \angle {MCB} = \varphi \) . The trigonometric form of\n\nCeva’s Theorem, applied twice to \( \bigtriangleup {ABC} \), with respect to \( N \) and \( M \), yields\n\n\[ \frac{\sin \angle {NBC}}{\sin \angle {NBA}} = \frac{\sin \left( {\gamma + \varphi }\right) }{\cos \alpha \sin \varphi } ... | Yes |
Problem 4.12.8. Let \( {ABC} \) be a triangle. The points \( D \) and \( E \) are the midpoints of the sides \( {BC} \) and \( {CA} \), respectively. The lines through an arbitrary point \( X \in {AB} \), that are parallel to \( {AD} \) and \( {BE} \), intersect \( {BC} \) and \( {CA} \) at the points \( N \) and \( M ... | Solution. Menelaus’ Theorem, applied to \( \bigtriangleup {MNC} \) and the line \( {EK} \) , yields\n\n\[ 1 = \frac{ME}{EC} \cdot \frac{CB}{BN} \cdot \frac{NK}{KM} = \frac{ME}{EA} \cdot \frac{2BD}{BN} \cdot \frac{NK}{KM} \]\n\n\[ = \frac{XB}{AB} \cdot \frac{2AB}{BX} \cdot \frac{NK}{KM} = \frac{2NK}{KM} \]\n\nwhich impl... | Yes |
Problem 4.12.9. Let \( {ABC} \) and \( {DEF} \) be similar triangles. Denote the midpoints of the segments \( {AD},{BE} \) and \( {CF} \) by \( M, N \) and \( P \), respectively. Prove that \( \bigtriangleup {MNP} \sim \bigtriangleup {ABC} \) . | Solution. We will use rotation of vectors. When a vector is multiplied by \( {e}^{i\varphi } \), it means that it is rotated by the angle \( \varphi \) counterclockwise. Let \( \overrightarrow{AB}.k.{e}^{i\varphi } = \overrightarrow{DE} \) . The given similarity implies \( \overrightarrow{BC}.k.{e}^{i\varphi } = \overr... | Yes |
Problem 4.12.10. Let \( \angle {AXD} \) and \( \angle {BXC} \) be right angles and let \( {AC} \cap {BD} = E \) and \( {AB} \cap {DC} = F \) . Prove that \( \angle {FXE} = {90}^{ \circ } \) . | Solution. The statement is a direct consequence of Problem 3.10. | No |
Problem 4.12.11. (Morley's Theorem) Let \( {ABC} \) be a triangle. Its angle trisectors intersect as shown in the figure. Prove that \( \bigtriangleup {QMN} \) is equilateral. | Solution. We will use the notations on the figure. Problem 4.12.12 gives \( {MK} = {KQ},{QP} = {PN} \) and \( {NL} = {LM} \) . Thus, we have\n\n\[ \angle {NQM} = {180}^{ \circ } - \angle {CQP} - \angle {PQN} - \angle {KQM} \]\n\n\[ = {180}^{ \circ } - \angle {CAQ} - \angle {ACQ} - \left( {{90}^{ \circ } - \frac{\angle ... | Yes |
Problem 4.12.12. Let \( {ABC} \) be a triangle. Its angle trisectors intersect as shown in the figure. Prove that \( {PQ} = {PN} \) . | Solution. We have that \( \angle {BAP} = \frac{2}{3}\alpha ,\angle {ABP} = \frac{2}{3}\beta \) and \( {AB} = c \) .\n\nLet \( M \) be the incenter of \( \bigtriangleup {ABP} \) and let \( {N}^{\prime } \) be a point on \( {BP} \), such that \( \angle {N}^{\prime }{MB} = {120}^{ \circ } - \frac{\beta }{3} - \frac{\gamma... | Yes |
Problem 5.1.1. Let \( {ABCD} \) be a parallelogram. Let \( E \) be the midpoint of \( {AB} \) and let \( F \) be the foot of the perpendicular from \( D \) to the line | Solution. Let \( {CE} \cap {DA} = P \) . Since \( {DC} = {2AE} \) and \( {DC}\parallel {AE} \), we have that \( {AE} \) is a midsegment in \( \bigtriangleup {PCD} \) . Hence, \( {AP} = {AD} \) and thus \( {AF} \) is the median to the hypotenuse in the right-angled \( \bigtriangleup {PFD} \) . We deduce that \( {AF} = {... | No |
Let \( {ABCD} \) be a parallelogram. The points \( E \) and \( F \) are chosen on the segments \( {DC} \) and \( {AD} \), respectively, so that \( {EC} = {AF} \) . Let \( {AE} \cap {CF} = P \) . Prove that \( {BP} \) is the angle bisector of \( \angle {ABC} \) . | Solution. Let \( {AE} \cap {BC} = Q \) . Then the intercept theorem yields\n\n\[ \frac{AP}{PQ} = \frac{AF}{CQ} = \frac{EC}{CQ} = \frac{AB}{BQ} \]\n\nNow, the angle bisector theorem yields that \( {BP} \) bisects \( \angle {ABC} \) . | Yes |
Let \( {ABCD} \) be a parallelogram. A point \( P \) is chosen inside of \( {ABCD} \), such that \( \angle {DAP} = \angle {DCP} \) . Prove that \( \angle {PBA} = \) \( \angle {PDA} \) . | Solution. We have\n\n\[ \angle {BAP} = \angle {BAD} - \angle {PAD} = \angle {BCD} - \angle {DCP} = \angle {BCP}. \]\n\nDenote \( \angle {ABC} = \angle {ADC} = \alpha ,\angle {PBA} = \beta \) and \( \angle {PDA} = {\beta }_{1} \) . The trigonometric form of Ceva’s Theorem, applied to the quadrilateral ABCD and the point... | Yes |
Let \( {ABCD} \) be a parallelogram with \( {AC} \cap {BD} = E \) . Let \( P \) be a point on \( {CD} \), such that \( \angle {EPC} = \angle {BCD} \) . Prove that \( {AP} = \) \( {BP} \) . | Solution. Let \( {PE} \cap {AB} = Q \) . The properties of the parallelogram yield \( {PE} = {EQ} \) . Hence, \( {QBPD} \) is also a parallelogram.\n\nMoreover, we have \( \angle {QAD} = \angle {BCD} = \angle {EPC} = \angle {PQA} \) . Hence, \( {AQPD} \) is an isosceles trapezoid and therefore \( {BP} = {DQ} = {AP} \) ... | Yes |
Problem 5.1.5. Let \( {ABCD} \) be a parallelogram. The circles \( {k}_{1}\left( {A;{AB}}\right) \) and \( {k}_{2}\left( {C;{CB}}\right) \) are constructed. Let \( P \in {k}_{1} \) be a point inside of \( {ABCD} \) , and let \( {BP} \cap {k}_{2} = \{ B, Q\} \) . Prove that \( {DP} = {DQ} \) . | Solution. We have \( {CD} = {AB} = {AP} \) and \( {CQ} = {CB} = {AD} \) . Moreover,\n\n\[ \angle {DAP} = \angle {DAB} - \angle {PAB} = {180}^{ \circ } - \angle {ABC} - {180}^{ \circ } + 2\angle {PBA} \]\n\n\[ = 2\angle {PBA} - \angle {ABC} = 2\left( {\angle {ABC} - \angle {QBC}}\right) - \angle {ABC} \]\n\n\[ = \angle ... | Yes |
Problem 5.1.6. Let \( {ABCD} \) be a parallelogram. The angle bisector of \( \angle {BAD} \) intersects the line \( {BC} \) and the segment \( {DC} \) at the points \( E \) and \( F \), respectively. Let \( O \) be the circumcenter of \( \bigtriangleup {EFC} \) . Prove that the quadrilateral \( {BDOC} \) is cyclic. | Solution. We have \( \angle {EFC} = \angle {DFA} = \angle {FAB} = \angle {FAD} = \angle {FEC} \) . Hence, \( {CF} = {CE} \) and \( {CD} = {BA} = {BE} \) . Since \( \bigtriangleup {EFC} \) is isosceles, we have \( \bigtriangleup {COF} \cong \bigtriangleup {COE} \) and \( \angle {OCD} = \angle {OEB} \) . Also, \( {OE} = ... | Yes |
Problem 5.1.7. Let \( {ABCD} \) be a parallelogram with \( {AC} \cap {BD} = M \) . Let the feet of the perpendiculars from \( B \) to the lines \( {AD},{AC} \) and \( {CD} \) be \( P, F \) and \( E \), respectively. Prove that the quadrilateral \( {PMFE} \) is cyclic. | Solution. Note that \( \angle {BPD} = \angle {BED} = {90}^{ \circ } \), so \( {BM} = {MD} = {MP} = \) \( {ME} \) . Moreover, we have\n\n\[ \angle {EFC} = \angle {EBC} = {90}^{ \circ } - \angle {ECB} = {90}^{ \circ } - \angle {BAP} = \angle {PBA} = \angle {PFA}. \]\n\nHence, \( {FM} \) is an external bisector in \( \big... | Yes |
Let \( {ABCD} \) be a rhombus. The point \( P \) lies on \( {BD} \) . The line \( {AP} \) intersects the segment \( {DC} \) at the point \( Q \) . The circumcircle of \( \bigtriangleup {ADQ} \) intersects \( {BD} \) for the second time at the point \( K \) . Prove that the quadrilateral \( {CQPK} \) is cyclic. | Solution. Since \( {ABCD} \) is a rhombus, we have \( {CD} = {DA} \) and \( \angle {ADP} = \) \( \angle {CDP} \) . Hence, \( \bigtriangleup {ADP} \cong \bigtriangleup {CDP} \) and \( \angle {QCP} = \angle {DCP} = \angle {DAP} = \) \( \angle {PKQ} \) . Therefore, the quadrilateral \( {CQPK} \) is cyclic. | Yes |
Problem 5.1.9. Let \( {ABCD} \) be a parallelogram. The point \( P \) lies on the segment \( {CD} \) . Let \( {O}_{1},{O}_{2},{O}_{3} \) be the circumcenters of \( \bigtriangleup {ADP},\bigtriangleup {BPC} \) and \( \bigtriangleup {ABP} \), respectively. Prove that the orthocenter of \( \bigtriangleup {O}_{1}{O}_{2}{O}... | Solution. Let \( M \) and \( N \) be the midpoints of \( {AP} \) and \( {BP} \), respectively.\n\nThen \( \angle {AM}{O}_{1} = {90}^{ \circ } \) and \( \angle {O}_{2}{NP} = {90}^{ \circ } \) . We have\n\n\[ \angle {O}_{3}{O}_{2}P = \angle N{O}_{2}P = \angle {BCP} = {180}^{ \circ } - \angle {ADP} = \angle M{O}_{1}P = \a... | Yes |
Problem 5.1.10. Let \( {OA},{OB} \) and \( {OC} \) be line segments. The points \( D, E \) and \( F \) are such that the quadrilaterals \( {AOBD},{BOCE} \) and \( {AOCF} \) are parallelograms. Prove that the circumcircles of \( \bigtriangleup {ABD},\bigtriangleup {BEC} \) and \( \bigtriangleup {ACF} \) have a common po... | Solution. Let \( M \) be the second intersection point of the circumcircles of \( \bigtriangleup {ABD} \) and \( \bigtriangleup {AFC} \) . It suffices to prove that the quadrilateral \( {BECM} \) is cyclic. Note that\n\n\[ \angle {BMC} = \angle {BMA} - \angle {CMA} = \angle {BDA} - \left( {{180}^{ \circ } - \angle {CFA... | Yes |
Problem 5.2.1. (Steiner’s Theorem) Let \( {ABCD}\left( {{AB}\parallel {CD}}\right) \) be a trapezoid and let \( M \) and \( N \) be the midpoints of \( {AB} \) and \( {CD} \), respectively. Denote \( {AC} \cap {BD} = F \) and \( {AD} \cap {BC} = E \) . Then the points \( M, F, N \) and \( E \) are collinear. | Solution: The fact that \( {AB}\parallel {CD} \) proves the existence of a homothety with center \( F \), such that \( C \rightarrow A \) and \( D \rightarrow B \) . Therefore, \( {CD} \rightarrow {AB} \) . Hence, the image of the midpoint of \( {CD} \) is the midpoint of \( {AB} \) .\n\nSo \( N \rightarrow M \) . Ther... | Yes |
Problem 5.2.2. Let \( {ABCD}\left( {{AB}\parallel {CD}}\right) \) be a circumscribed trapezoid. Let \( {AD} \cap {BC} = E \) . The point \( P \) lies on the segment \( {AB} \) and \( {EP} \cap {CD} = Q \) . The line, parallel to \( {AB} \) and tangent to the incircle of \( \bigtriangleup {APE} \), intersects \( {EP} \)... | Solution. The problem is analogous to Problem 5.4.1. Indeed, choose the point \( B \) in the configuration of Problem 5.4.1 to be infinite. This choice preserves the circles in the diagram. | No |
Problem 5.2.3. Let \( {ABCD}\left( {{AB}\parallel {CD}}\right) \) be a trapezoid. The diagonals \( {AC} \) and \( {BD} \) intersect at \( M \) . The circles \( {k}_{1} \) and \( {k}_{2} \) with diameters \( {AD} \) and \( {BC} \), respectively, do not intersect each other. Prove that \( M \) lies on the radical axis of... | Solution. Denote the feet of the perpendiculars from \( C \) to \( {BD} \) and from \( D \) to \( {AC} \) by \( {H}_{2} \) and \( {H}_{1} \), respectively. Then \( {H}_{2} \in {k}_{2} \) and \( {H}_{1} \in {k}_{1} \) . Moreover, the quadrilateral \( {CD}{H}_{1}{H}_{2} \) is cyclic. Now we have\n\n\[ \angle {H}_{1}{AB} ... | Yes |
Problem 5.2.4. Let \( {ABCD}\left( {{AB}\parallel {CD}}\right) \) be a trapezoid. The diagonals \( {AC} \) and \( {BD} \) intersect at \( M \) . The circles \( {k}_{1} \) and \( {k}_{2} \) with diameters \( {AD} \) and \( {BC} \), respectively, intersect each other. Prove that \( M \) lies on the radical axis of \( {k}... | Solution. The solution is completely analogous to that of Problem 5.2.3. | No |
Problem 5.2.5. A trapezoid \( {ABCD}\left( {{AB}\parallel {CD}}\right) \) is inscribed in a circle with center \( O \) and radius \( R \) . Let \( M \) be the midpoint of \( {AB} \) . Let one of the intersection points of the circles with diameters \( {AB} \) and \( {CD} \) be \( K \) . The point \( H \) is the foot of... | Solution. Since the trapezoid is inscribed, it is also isosceles. Let \( N \) be the midpoint of \( {CD} \) and let \( {KH} \cap {AB} = L \) .\n\nConsidering the power of the point \( H \), we get \( {R}^{2} - O{H}^{2} = {DH}.{CH} = \) \( K{H}^{2} \), where the last equality follows from \( \bigtriangleup {CHK} \sim \b... | Yes |
Problem 5.2.6. Let \( {ABCD}\left( {{AB}\parallel {CD}}\right) \) be a trapezoid with \( {AC} \cap \) \( {BD} = M \) . The circumcircle of \( \bigtriangleup {AMD} \) intersects the segments \( {AB} \) and \( {CD} \) for the second time at the points \( F \) and \( K \), respectively. The circumcircle of \( \bigtriangle... | Solution. The quadrilateral NPCB is cyclic, and so it is an isosceles trapezoid. Therefore, \( \overset{⏜}{PN} = \overset{⏜}{BC} \) . We have\n\n\[ \angle {NEP} = {180}^{ \circ } - \angle {BMC} = \angle {AMB} = \angle {AME} + \angle {BME} \]\n\n\[ = \angle {EFN} + \angle {ENF} = {180}^{ \circ } - \angle {FEN}\text{. } ... | Yes |
Problem 5.2.7. Let \( {ABCD}\left( {{AB}\parallel {CD}}\right) \) be a trapezoid. The points \( P \) and \( Q \) lie on the segments \( {AC} \) and \( {BD} \), respectively. If \( \angle {BPD} = \angle {AQC} \) , prove that the quadrilateral \( {PQCD} \) is cyclic. | Solution. Let \( {Q}^{\prime } \in {BD} \) be a point such that \( {AP}{Q}^{\prime }B \) is cyclic and \( {Q}^{\prime } ≢ \) \( B \) . We have \( \angle P{Q}^{\prime }D = \angle {PAB} = \angle {PCD} \) and hence \( P{Q}^{\prime }{CD} \) is also cyclic. Therefore, \( \angle {BPD} = \angle {BPC} + \angle {DPC} = \angle A... | Yes |
Let \( {ABCD}\left( {{AB}\parallel {CD}}\right) \) be a trapezoid. Let \( M \) and \( N \) be the midpoints of \( {AC} \) and \( {BD} \), respectively. If \( \angle {NAB} = \angle {CAD} \) , prove that \( \angle {ADM} = \angle {BDC} \) . | We have\n\n\n\n\[ \frac{\sin \angle {BAC}}{\sin \angle {CAD}} = \frac{\sin \angle {DAN}}{\sin \angle {NAB}} \]\n\n\[ \Rightarrow \frac{BC}{CD} \cdot \frac{\sin \angle {ABC}}{\sin \angle {ADC}} = \frac{AB}{AD} \]\n\n\... | Yes |
Problem 5.2.9. Let \( {ABCD}\left( {{AB}\parallel {CD}}\right) \) be a trapezoid. Let \( {AC} \cap \) \( {BD} = F \) and \( {AD} \cap {BC} = E \) . The circumcircles \( {k}_{1} \) and \( {k}_{2} \) of \( \bigtriangleup {AFD} \) and \( \bigtriangleup {BCF} \), respectively, intersect for the second time at the point \( ... | Solution. Consider the composition \( \varphi \) of inversion with center \( E \) and radius \( \sqrt{{EC}.{EA}} \), and symmetry with respect to the angle bisector of \( \angle {AEB} \) .\n\nSince \( {AB}\parallel {CD} \), the equality \( {EC}.{EA} = {ED}.{EB} \) holds. Therefore, \( \varphi \left( A\right) = C \) and... | Yes |
Problem 5.2.10. Let \( {ABCD}\left( {{AB}\parallel {CD}}\right) \) be a trapezoid. Let \( {AD} \cap \) \( {BC} = E \) . The circumcircles \( {k}_{1} \) and \( {k}_{2} \) of \( \bigtriangleup {BCD} \) and \( \bigtriangleup {ABD} \) have centers \( {O}_{1} \) and \( {O}_{2} \), respectively. Prove that \( \angle {AE}{O}_... | Solution. As in the previous problem, consider the composition \( \varphi \) of inversion with center \( E \) and radius \( \sqrt{{EC}.{EA}} \), and symmetry with respect to the angle bisector of \( \angle {AEB} \) . \n\nSince \( {AB}\parallel {CD} \), the equality \( {EC}.{EA} = {ED}.{EB} \) holds. Hence, \( \varphi \... | Yes |
Problem 5.3.1. Let \( {ABCD} \) be a square and let \( E \) be a point on the segment \( {BC} \) . The square \( {BKFE} \) is constructed, such that it is external to \( {ABCD} \) . Let \( {AE} \cap {CK} = M \) and \( {KE} \cap {AC} = N \) . Prove that the points \( D, N, M \) and \( F \) are collinear. | Solution. We have \( {AB} = {BC} \) and \( {BE} = {BK} \) . Hence, \( \bigtriangleup {ABE} \cong \) \( \bigtriangleup {CBK} \) . Thus, \( \angle {BCK} = \angle {BAM} \) . This implies that the quadrilateral \( {ABMC} \) is cyclic, which yields \( \angle {AMK} = \angle {CBA} = {90}^{ \circ } \) .\n\nAlso, \( \angle {EFK... | Yes |
Problem 5.3.2. Let \( {ABCD} \) be a square. Its incircle \( k \) touches the side \( {AD} \) at the point \( M \) . The point \( P \) lies on the segment \( {AB} \) and satisfies \( \angle {PMA} = \alpha > {45}^{ \circ } \) . The point \( Q \) lies on the segment \( {BC} \), such that \( {QD}\parallel {PM} \) . Prove ... | Solution. Without loss of generality, let\n\n\n\n\( {AB} = 2 \) . Then \( {AM} = {MD} = 1 \) .\n\nWe have \( \angle {DQC} = \angle {ADQ} = \alpha \) and \( {AP} = \operatorname{tg}\alpha \) . Hence, \( \operatorname{... | Yes |
Problem 5.3.3. Let \( {ABCD} \) be a square and let \( P \) be a point inside of it. Prove that the perpendiculars from \( A \) to \( {DP} \), from \( B \) to \( {AP} \), from \( C \) to \( {BP} \), and from \( D \) to \( {CP} \) are concurrent. | Solution. Consider a \( + {90}^{ \circ } \) rotation with center \( O \) (the center of the square). Then \( A \rightarrow B, B \rightarrow C, C \rightarrow D \) and \( D \rightarrow A \) . Let \( P \rightarrow Q \) . Then we have the following congruences:\n\n\[ \bigtriangleup {ABQ} \cong \bigtriangleup {DAP} \]\n\[ \... | Yes |
Let \( {BAD} \) be a triangle. The circles \( {k}_{1} \) and \( {k}_{2} \) , are inscribed in \( \angle {BAD} \), such that they both touch the segments \( {AB} \) and \( {AD} \) and not their extensions. The tangent lines through \( B \) and \( D \) to \( {k}_{1} \) (different from \( {AB} \) and \( {AD} \) ) intersec... | Solution. Let \( {BN} \cap {AD} = P \) and let \( {DN} \cap {AB} = M \) . The quadrilateral \( {ABCD} \) is circumscribed and thus \( {AB} + {CD} = {AD} + {BC} \), i.e. \( {BC} - {CD} = \) \( {BA} - {AD} \) . The quadrilateral \( {AMNP} \) is also circumscribed. Thus, \( {BA} - {AD} = \) \( {BN} - {ND} \), which gives ... | Yes |
Problem 5.4.2. Let \( {ABCD} \) be a circumscribed quadrilateral. The circles \( {k}_{1} \) and \( {k}_{2} \) are inscribed in \( \angle {BAD} \) and \( \angle {BCD} \), respectively. The tangent lines through \( B \) and \( D \) to \( {k}_{2} \) (different from \( {DC} \) and \( {BC} \) ) intersect each other at the p... | Solution. Note that Problem 5.4.1, applied to the quadrilateral ABCD, its incircle and \( {k}_{2} \), yields that the quadrilateral \( {ABND} \) is circumscribed.\n\nThe same problem, applied to the quadrilateral ABND, its incircle and \( {k}_{1} \), yields that the quadrilateral \( {QBND} \) is circumscribed. | No |
Problem 5.4.3. Let \( {ABCD} \) be a circumscribed quadrilateral. The circles \( {k}_{1} \) and \( {k}_{2} \) are inscribed in \( \angle {BAD} \) and \( \angle {ABC} \), respectively. The tangent lines through \( C \) and \( D \) to \( {k}_{2} \) and \( {k}_{1} \), different from \( {CB} \) and \( {AD} \) , intersect \... | Solution. Let the line \( {RS} \n\n\n\nintersect the line \( {AB} \) and the segments \( {AD} \) and \( {BC} \) at the points \( P, Q \) and \( T \), respectively.\n\nSince the quadrilateral \( {AMRQ} \) is circumscr... | Yes |
Problem 5.4.4. Let \( {ABCD} \) be a circumscribed quadrilateral. The three circles \( {k}_{1},{k}_{3} \) and \( {k}_{2} \) are inscribed in \( \angle {BAD},\angle {ABC} \) and \( \angle {BCD} \) , respectively. The tangent lines through \( D \) to \( {k}_{1} \) and \( {k}_{2} \) (different from \( {AD} \) and \( {DC} ... | Solution. Using the notations on the figure, we have\n\n\[ {DU} + {UV} = {DC} + {CV} \]\n\n(due to the circumscribed quadrilateral \( {UTCL} \) ) and\n\n\[ {RV} + {MB} = {RM} + {BV} \]\n\n(due to the circumscribed quadrilateral \( {QBSR} \) ).\n\nNow we subtract the second equality from the first to get\n\n\[ {DU} - {U... | Yes |
Problem 5.4.5. Let \( {ABCD} \) be a circumscribed quadrilateral. The circles \( {k}_{1},{k}_{2},{k}_{3},{k}_{4} \) are inscribed in \( \angle {BAD},\angle {ABC},\angle {BCD} \) and \( \angle {CDA} \) , respectively. Denote the common external tangent lines of \( {k}_{1} \) and \( {k}_{2} \), of \( {k}_{2} \) and \( {k... | Solution. Using the notations on the figure, we have\n\n\[ \n{MK} + {IJ} = {MS} - {KS} + {IG} - {JG} \]\n\n\[ \n= \left( {{SD} + {DT} - {MT}}\right) - \left( {{KU} + {CS} - {UC}}\right) + \left( {{AG} + {IT} - {AT}}\right) \]\n\n\[ \n- \left( {{GB} + {BU} - {UJ}}\right) = \left( {{SD} - {SC}}\right) + \left( {{DT} - {A... | Yes |
Problem 5.4.6. Let \( {ABCD} \) be a circumscribed quadrilateral. Prove that its incenter \( I \) and the midpoints \( M \) and \( N \) of the segments \( {AC} \) and \( {BD} \), respectively, are collinear. | Solution. It suffices to prove that the oriented area of \( \bigtriangleup {MIN} \) satisfies \( \left\lbrack {MIN}\right\rbrack = 0 \) . We have\n\n\[ \left\lbrack {MIN}\right\rbrack = \frac{\left\lbrack {MIB}\right\rbrack + \left\lbrack {MID}\right\rbrack }{2} = \frac{\frac{\left\lbrack {AIB}\right\rbrack + \left\lbr... | Yes |
Problem 5.4.7. Let \( {ABCD} \) be a circumscribed quadrilateral with incircle \( k \) . The circles \( {k}_{1}\left( {I}_{1}\right) ,{k}_{2}\left( {I}_{2}\right) ,{k}_{3}\left( {I}_{3}\right) ,{k}_{4}\left( {I}_{4}\right) \) are the incircles of \( \bigtriangleup {ARU},\bigtriangleup {RBS},\bigtriangleup {SCT} \) and ... | Solution. Let \( K = {t}_{12} \cap {AB}, L = {t}_{34} \cap {CD}, G = {t}_{14} \cap {AD}, H = \) \( {t}_{23} \cap {BC}, F = {AB} \cap {CD} \) and \( E = {AD} \cap {BC} \) .\n\n\n\nDenote the angles of \( {ABCD} \) by \( \alpha ,\beta ,\gamma \) and \( \delta \) . It is easy to see that the quadrilateral \( {I}_{1}{I}_{2... | Yes |
Problem 5.4.8. Let \( {ABCD} \) be a circumscribed quadrilateral with incircle \( k \) . The circle \( {k}_{1} \) is inscribed in \( \angle {ABC} \) and intersects the segment \( {AC} \) at the points \( M \) and \( N \) . Prove that the circle, which passes through \( M \) and \( N \) and touches \( {CD} \), also touc... | Solution. We will prove the following lemma:\n\nLemma. An angle \( \angle {CDA} \) and a line \( {MN} \), not containing \( D \), are given. The circle \( {k}_{1} \) passes through \( M \) and \( N \) and touches \( {DA} \) at the point \( Y \), and the circle \( {k}_{2} \) passes through \( M \) and \( N \) and touche... | Yes |
Let \( {ABCD} \) be a circumscribed quadrilateral with incircle \( k \) with center \( I \). The feet of the perpendiculars from \( B \) and \( D \) to the angle bisector of \( \angle {BAD} \) are \( M \) and \( P \), respectively, and the feet of the perpendiculars from \( B \) and \( D \) to the angle bisector of \( ... | It suffices to prove that \( \angle {MNI} = \angle {MPQ} \). Note that \[ \angle {MNI} = \angle {MBI} = {90}^{ \circ } - \angle {MIB} \] \[ = {90}^{ \circ } - \left( {{180}^{ \circ } - \angle {CID}}\right) \] \[ = \angle {CID} - {90}^{ \circ } = \angle {QDI} \] \[ = \angle {QPI} \] and the desired statement follows. | Yes |
Problem 5.4.11. Let \( {ABCD} \) be a circumscribed quadrilateral with incircle \( k \) with center \( I \). Let \( {ABCD} \) be also inscribed in a circle \( {k}_{1} \). The points \( T \) and \( J \) are the incenters of \( \bigtriangleup {ABC} \) and \( \bigtriangleup {ADC} \), respectively. Prove that the quadrilat... | Solution. Let the incircle\n\n\n\n\( {k}_{2} \) of \( \bigtriangleup {ADC} \) touch \( {AC} \) at the point \( X \).\n\nThen \( {CX} - {XA} = \)\n\n\[ = {CD} - {DA} = {CB} - {BA}\text{.}\]\n\nHence, \( X \) is the po... | No |
Problem 5.4.12. Let \( {ABCD} \) be a circumscribed quadrilateral with incircle \( k \) with center \( I \) . Let \( {ABCD} \) be also inscribed in a circle \( {k}_{1} \) . Let \( k \) touch \( {AB},{BC},{CD} \) and \( {DA} \) at the points \( M, N, P \) and \( Q \), respectively. Prove that \( {MP} \bot {NQ} \) . | Solution. We have\n\n\n\n\[ \n\angle \left( {{MP},{NQ}}\right) = \n\] \n\n\[ \n= \frac{\overset{⏜}{PN} + \overset{⏜}{QM}}{2} \n\] \n\n\[ \n= \frac{\angle {PIN} + \angle {QIM}}{2} \n\] \n\n\[ \n= \frac{{360}^{ \circ }... | Yes |
Problem 5.4.13. Let \( {ABCD} \) be a circumscribed quadrilateral with incircle \( k \) with center \( I \) . The line \( {AC} \) intersects \( k \) at \( Y \) and \( Z \) . The point \( X \) is the midpoint of \( {YZ} \) . Prove that \( \angle {BXC} = \angle {DXC} \) . | Solution. Let \( {IX} \cap {BD} = S \) . Since \( \angle {IXC} = {90}^{ \circ } \), it suffices to prove that \( \left( {D, L, B, S}\right) = - 1 \) . Let \( {QM} \cap {PN} = {S}^{\prime } \) .\n\nSince the polar lines of \( D, B \) and \( {S}^{\prime } \) with respect to \( k \) pass through the intersection point of ... | Yes |
Problem 5.4.14. Let \( {ABCD} \) be a circumscribed quadrilateral with incircle \( k \) with center \( I \). The line through \( A \), perpendicular to \( {AD} \), intersects the angle bisector of \( \angle {ADC} \) at the point \( T \), and the line through \( A \), perpendicular to \( {AB} \), intersects the angle bi... | Solution. Let \( k \) touch \( {AD} \) and \( {AB} \) at the points \( H \) and \( E \), respectively. It suffices to prove that the line connecting the pole of \( {ST} \) and \( I \) is parallel to \( {AC} \). The pole of the line \( {AT} \) lies on the line through \( I \), parallel to \( {AD} \), and at the same tim... | Yes |
Problem 5.4.15. Let \( k \) be a circle inscribed in the angle \( \angle {ABC} \) . Let \( k \) touch \( {AB} \) and \( {BC} \) at the points \( M \) and \( P \), respectively. A line through \( B \) intersects \( k \) at the points \( N \) and \( Q \), such that \( N \) lies between \( B \) and \( Q \) . The point \( ... | Solution. Since \( M \) and \( P \) lie on the polar line of \( B \) with respect to \( k \), then the tangent lines at the points \( N \) and \( Q \) intersect on \( {MP} \) . Therefore, \( {MP} \) is a symmedian in \( \bigtriangleup {QPN} \) and in \( \bigtriangleup {QMN} \) . Problem 4.4.1 yields \[ \angle {XPQ} = \... | Yes |
Problem 5.4.16. Let \( {ABCD} \) be a circumscribed quadrilateral with \( {BC} \cap {AD} = F \) and \( {BA} \cap {CD} = E \) . Let \( N \in {BC} \) and \( M \in {AB} \) . The line \( {FM} \) intersects the lines \( {EN} \) and \( {DC} \) at the points \( X \) and \( P \) , respectively. Let \( {EX} \cap {AD} = Q \) . I... | Solution. We have \( {FC} + {CE} = {FA} + {AE} = {FX} + {XE} \) . Hence, the quadrilateral \( {CNXP} \) is circumscribed. | No |
Problem 5.4.17. Let \( {ABCD} \) be a circumscribed quadrilateral with incircle \( k \) . Let \( {AD} \cap {BC} = E \) and \( {AB} \cap {CD} = F \) . The circles \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) are inscribed in \( \angle {ABC},\angle {BCD} \) and \( \angle {CDA} \), respectively. One of the common internal tang... | Solution. Problem 5.4.16, applied to the quadrilateral ABCD and the lines \( {FM} \) and \( {EG} \), yields that the quadrilateral \( {MSLD} \) is circumscribed.\n\nThe same problem, applied to the quadrilateral \( {NQKD} \) and the lines \( {EP} \) and \( {FM} \), yields that the quadrilateral \( {PQRS} \) is circumsc... | Yes |
Let \( {ABCD} \) be a convex quadrilateral. The bisectors \( {l}_{a},{l}_{b},{l}_{c} \) and \( {l}_{d} \) of the external angles of \( {ABCD} \) form the quadrilateral \( {EFGH} \), where \( {l}_{a} \cap {l}_{b} = E,{l}_{b} \cap {l}_{c} = F,{l}_{c} \cap {l}_{d} = G \) and \( {l}_{d} \cap {l}_{a} = H \) . The perpendicu... | Clearly, \( \angle {KHA} = \angle {KEA} \) and thus \( \bigtriangleup {KHE} \) is isosceles, which implies that the angle bisector of \( \angle {LKN} \) coincides with the perpendicular bisector of \( {HE} \) . We can make a similar argument for the other three angles.\n\nWe come to the conclusion that the four bisecto... | Yes |
Problem 5.5.1. Let \( {ABCD} \) be a cyclic quadrilateral with \( {AB} \cap {CD} = \) \( E \) and \( {AD} \cap {BC} = F \) . The points \( K, H \in {CD} \) and \( L, G \in {AB} \) are chosen, such that the quadrilateral \( {LGHK} \) is cyclic and the triples of points \( F \) , \( K, L \) and \( F, H, G \) are collinea... | Solution. We have\n\n\( \begin{aligned} \angle {SVU} & = \angle {VLG} + \angle {MRV} - \angle {LAR} \\ & = \angle {THC} + \angle {TNC} - \angle {BCF} = \angle {STH}, \end{aligned} \) as desired. | Yes |
Let \( {ABCD} \) be a cyclic quadrilateral with circumcircle \( k \) . The tangent lines at the points \( D \) and \( C \) intersect at the point \( E \) . The lines through \( E \), parallel to \( {AC} \) and \( {BD} \), intersect \( {BC} \) and \( {AD} \) at the points \( F \) and \( G \), respectively. Prove that th... | We have\n\n\n\n\[ \angle {ECF} = \frac{\overset{⏜}{BC}}{2} = \angle {BAC} \]\n\nand \( \angle {EFC} = \angle {ACB} \), which gives \( \bigtriangleup {ECF} \sim \bigtriangleup {BAC} \) . Analogously, we deduce that \(... | Yes |
Let \( {ABCD} \) be a cyclic quadrilateral with circumcircle \( k \) . The lines, symmetric to \( {BD} \) with respect to the angle bisectors of \( \angle {ADC} \) and \( \angle {ABC} \), intersect at the point \( F \) . Prove that \( {FA} = {FC} \) . | Let \( {DF} \) intersect \( k \) at the point \( G \) . Then \( \overset{⏜}{AG} = \overset{⏜}{BC} \) . On the other hand, \( {AGBC} \) is an isosceles trapezoid. But\n\n\[ \angle {DGB} = \angle {DGC} + \angle {CGB} = \angle {CBD} + \angle {CDB} = \angle {FBA} + \angle {ADG} = \angle {FBG}. \]\n\nHence, \( \bigtriangleu... | Yes |
Let \( {ABCD} \) be a cyclic quadrilateral with circumcircle \( k \) . Let \( M \) and \( N \) be the midpoints of \( {AC} \) and \( {BD} \), respectively. If \( \angle {ABM} = \angle {CBD} \), prove that \( \angle {BCA} = \angle {NCD} \) . | We have that \( {BD} \) is a symmedian in \( \bigtriangleup {ABC} \) and hence it passes through the intersection point \( X \) of the tangent lines at \( A \) and \( C \) to \( k \) . Thus, the quadrilateral \( {ABCD} \) is harmonic.\n\nThis implies that the tangent lines at \( B \) and \( D \) to \( k \), and the lin... | No |
Let \( {ABCD} \) be a cyclic quadrilateral with circumcircle \( k \) with center \( O \) . Let \( {AC} \cap {BD} = E \) . The external angle bisectors through \( A, B, C \) and \( D \) form the quadrilateral \( {FKLP} \), as shown in the figure. Prove that \( {FKLP} \) is cyclic and that its center \( {O}_{1} \) lies o... | Solution. We have\n\n\[ \angle {DLC} + \angle {AFB} = {360}^{ \circ } - \angle {LCD} - \angle {LDC} - \angle {FAB} - \angle {FBA} \]\n\n\[ = {360}^{ \circ } - \angle {KCB} - \angle {PDA} - \angle {DAP} - \angle {KBC} \]\n\n\[ = \angle {APD} + \angle {BKC} \]\n\n\[ \Rightarrow \angle {DLC} + \angle {AFB} = \angle {APD} ... | Yes |
Problem 5.5.6. (Ptolemy's Theorem) If the quadrilateral ABCD is cyclic, prove that \( {AB} \cdot {CD} + {AD} \cdot {BC} = {AC} \cdot {BD} \) . | Solution. Set \( {AB} = a \) ,\n\n\n\n\( {DC} = b,{CB} = c,{DA} = d \) , \( {AC} = e \) and \( {BD} = f. \)\n\nConsider an inversion with center \( C \) and an arbitrary radius. Let the images of \( D, A \) and \( B ... | Yes |
Problem 5.5.8. Let \( {ABCD} \) be a cyclic quadrilateral with circumcircle \( k \) . Its diagonals intersect at the point \( X \) . The incenter of \( \bigtriangleup {XBC} \) is \( I \), and the center of the circle, external to \( {ABCD} \), that touches the lines \( {AB} \) and \( {CD} \) and the segment \( {BC} \) ... | Solution. Denote the incenters of \( \bigtriangleup {ABC} \) and \( \bigtriangleup {DBC} \) by \( K \) and \( L \) , respectively. Let the excircles of these triangles, tangent to the segment \( {BC} \), have centers \( S \) and \( T \), respectively.\n\nThe trillium theorem (Problem 4.6.1) and its analog for excircles... | Yes |
Problem 5.5.9. Let \( {ABCD} \) be a cyclic quadrilateral with circumcircle \( k \) . Its diagonals intersect at the point \( K \) . The incenters of \( \bigtriangleup {ABK} \) , \( \bigtriangleup {BCK},\bigtriangleup {CDK} \) and \( \bigtriangleup {DAK} \) are \( G, H, E \) and \( F \), respectively. The midpoints of ... | Solution. Let \( O \) be the center of \( k \) . It suffices to prove that the lines \( G{G}^{\prime }, H{H}^{\prime } \) and \( {OK} \) are concurrent.\n\nDesargues’ Theorem (Problem 7.1), applied to \( \bigtriangleup O{G}^{\prime }{H}^{\prime } \) and \( \bigtriangleup {KGH} \) , yields that it suffices to prove that... | Yes |
Let \( {ABCD} \) be a cyclic quadrilateral with circumcircle \( k \) . The incenters of \( \bigtriangleup {ABC},\bigtriangleup {BCD},\bigtriangleup {CDA} \) and \( \bigtriangleup {DAB} \) are \( F, H \) , \( E \) and \( G \) , respectively. Prove that the quadrilateral \( {EGFH} \) is a rectangle. | Solution. We have \( \angle {AGB} = {90}^{ \circ } + \frac{\angle {ADB}}{2} = {90}^{ \circ } + \frac{\angle {ACB}}{2} = \angle {AFB} \) . Hence, the quadrilateral ABFG is cyclic. Analogously, the quadrilateral \( {AGED} \) is cyclic. Then \[ \angle {FGE} = {360}^{ \circ } - \angle {AGF} - \angle {AGE} \] \[ = {360}^{ \... | Yes |
Let \( {ABCD} \) be a cyclic quadrilateral with circumcenter \( O \) and let \( {AD} \cap {BC} = E \) . The circumcircles of \( \bigtriangleup {ABE} \) and \( \bigtriangleup {CDE} \) intersect for the second time at the point \( H \) . Prove that \( \angle {EHO} = {90}^{ \circ } \) . | Solution. Let \( {AB} \cap {CD} = F,{AC} \cap {BD} = P \) and \( {OP} \cap {EF} = {H}^{\prime } \) . Brocard’s Theorem (Problem 4.8.25) yields that \( {OP} \bot {EF} \) . Thus, \( O{H}^{\prime } \bot {EF} \) . Let \( K \) and \( L \) be points such that \( {AK} \bot {AB},{DK} \bot {DC} \) , \( {LB} \bot {AB} \) and \( ... | Yes |
Problem 5.6.2. Let \( {ABDC} \) be a cyclic quadrilateral with circumcenter \( O \) and let \( {AD} \cap {BC} = E \) . The circumcircles of \( \bigtriangleup {ABE} \) and \( \bigtriangleup {CDE} \) intersect for the second time at the point \( H \) . Prove that \( \angle {EHO} = {90}^{ \circ } \). | Solution. This problem is essentially the same as Problem 5.6.1 but for the quadrilateral \( {ABDC} \) instead of \( {ABCD} \) . The solution is completely analogous. | No |
Problem 5.6.3. Let \( {ABCD} \) be a cyclic quadrilateral with circumcenter \( O \) and let \( {AB} \cap {CD} = F \) . The circumcircles of \( \bigtriangleup {ADO} \) and \( \bigtriangleup {BCO} \) intersect for the second time at the point \( Q \) . Prove that \( \angle {FQO} = {90}^{ \circ } \) . | Solution. Let us use the results and notations of Problem 5.6.1. Let \( {FP} \cap {EO} = {Q}^{\prime } \) . Brocard’s Theorem (Problem 4.8.24) yields \( {FP} \bot {EO} \) . Therefore, the quadrilateral \( {OFH}{Q}^{\prime } \) is cyclic. On the other hand, the quadrilateral \( {BCHF} \) is also cyclic. Hence, \[ E{Q}^{... | Yes |
Problem 5.6.4. Let \( {ADBC} \) be a cyclic quadrilateral with circumcenter \( O \) and let \( {AB} \cap {CD} = F \) . The circumcircles of \( \bigtriangleup {ADO} \) and \( \bigtriangleup {BCO} \) intersect for the second time at the point \( Q \) . Prove that \( \angle {FQO} = {90}^{ \circ } \) . | Solution. This problem is essentially the same as Problem 5.6.3 but for the quadrilateral \( {ADBC} \) instead of \( {ABCD} \) . The solution is completely analogous. | No |
Problem 5.6.5. Let \( {ABDC} \) be a cyclic quadrilateral with circumcenter \( O \) and let \( {AB} \cap {CD} = F \) . The circumcircles of \( \bigtriangleup {ADO} \) and \( \bigtriangleup {BCO} \) intersect for the second time at the point \( Q \) . Prove that \( \angle {FQO} = {90}^{ \circ } \) . | Solution. This problem is essentially the same as Problem 5.6.3 but for the quadrilateral \( {ABDC} \) instead of \( {ABCD} \) . The solution is completely analogous. | No |
Let \( {ABCD} \) be a cyclic quadrilateral with circumcircle \( k \) . Let \( {AC} \cap {BD} = P \) . The line \( l \) passes through \( P \) and intersects the circumcircles of \( \bigtriangleup {APD} \) and \( \bigtriangleup {BCP} \) for the second time at the points \( E \) and \( F \), respectively ( \( E \) and \(... | Set \( \angle {DAP} = \psi ,\angle {DAM} = \delta \) and \( \angle {CBN} = \varphi \) .\n\nTherefore, \( \angle {PED} = \psi = \angle {DBC} = \angle {PFC},\angle {FCN} = \angle {CNM} - \psi = \) \( \angle {CAM} - \psi = \angle {DAM} = \delta \) and \( \angle {EDM} = \angle {DMN} - \psi = \angle {DBN} - \psi = \) \( \an... | Yes |
Problem 5.6.7. Let \( {ACBD} \) be a cyclic quadrilateral with circumcircle \( k \) . Let \( {AC} \cap {BD} = P \) . The line \( l \) passes through \( P \) and intersects the circumcircles of \( \bigtriangleup {APD} \) and \( \bigtriangleup {BCP} \) for the second time at the points \( E \) and \( F \), respectively (... | Solution. This problem is essentially the same as Problem 5.6.6 but for a quadrilateral \( {ACBD} \) instead of \( {ABCD} \), and points \( E \) and \( F \) lying inside of \( k \) . The solution is completely analogous. | No |
Problem 5.6.8. Let \( {ACDB} \) be a cyclic quadrilateral with circumcircle \( k \) . Let \( {AC} \cap {BD} = P \) . The line \( l \) passes through \( P \) and intersects the circumcircles of \( \bigtriangleup {APD} \) and \( \bigtriangleup {BCP} \) for the second time at the points \( E \) and \( F \), respectively (... | Solution. This problem is essentially the same as Problem 5.6.6 but for a quadrilateral \( {ACDB} \) instead of \( {ABCD} \), and points \( E \) and \( F \) lying inside of \( k \) . The solution is completely analogous. | No |
Problem 5.6.9. Let \( {ABC} \) be a triangle. The points \( E \) and \( F \) lie on the segments \( {AC} \) and \( {BC} \), respectively. Denote the circumcircles of \( \bigtriangleup {ABC},\bigtriangleup {ACF} \) and \( \bigtriangleup {BCE} \) by \( k,{k}_{1} \) and \( {k}_{2} \), respectively. Let \( {EF} \cap k = \)... | Solution. Set \( {NL} = a,{FN} = b,{EF} = c,{ME} = d \) and \( {KM} = e \) . Considering the power of the point \( F \) with respect to \( k \) and \( {k}_{2} \), we get\n\n\[ c\left( {b + a}\right) = {CF} \cdot {BF} = b\left( {c + d}\right) \Rightarrow a = \frac{bd}{c}. \]\n\nConsidering the power of the point \( E \)... | Yes |
Problem 5.6.10. Let \( {ABCD} \) be a cyclic quadrilateral with circumcircle \( k \) . The points \( {O}_{1},{O}_{2},{O}_{3} \) and \( {O}_{4} \) are the midpoints of the small arcs \( \overset{⏜}{AB},\overset{⏜}{BC},\overset{⏜}{CD} \) and \( \overset{⏜}{DA} \), respectively. For the circles \( {k}_{1}\left( {{O}_{1};{... | Solution. The trillium theorem (Problem 4.6.1), applied to \( \bigtriangleup {BCD} \) , yields that if \( I \) is its incenter, then \( I{O}_{3} = {O}_{3}C \) and \( I{O}_{2} = {O}_{2}C \) . But \( N{O}_{3} = {O}_{3}C \) and \( N{O}_{2} = {O}_{2}C \) . Thus, \( N \equiv I \) .\n\nAnalogously, the points \( M, Q \) and ... | Yes |
Problem 5.6.11. Let \( {ABCD} \) be a cyclic quadrilateral with circumcircle \( k \) . Let \( {AD} \cap {BC} = E \) and \( {AC} \cap {BD} = P \) . The point \( M \) is the midpoint of \( {EP} \) . The circumcircle of \( \bigtriangleup {BPE} \) intersects \( k \) at the points \( B \) and \( X \) . The circumcircle of \... | Solution. We will prove that \( \frac{\sin \angle {EAM}}{\sin \angle {PAM}} = \frac{\sin \angle {EAX}}{\sin \angle {PAX}} \) . This will imply \( \cot \angle {EAM} = \cot \angle {EAX} \) and, consequently, \( \angle {EAM} = \angle {EAX} \), which will give the collinearity of \( A, X \) and \( M \) . We have \[ \frac{\... | Yes |
Problem 5.6.12. Let \( {CADB} \) be a cyclic quadrilateral with circumcircle \( k \) . Let \( {AD} \cap {BC} = E \) and \( {AC} \cap {BD} = P \) . The point \( M \) is the midpoint of \( {EP} \) . The circumcircle of \( \bigtriangleup {BPE} \) intersects \( k \) at the points \( B \) and \( X \) . The circumcircle of \... | Solution. This problem is essentially the same as Problem 5.6.11 but for the quadrilateral \( {CADB} \) instead of \( {ABCD} \) . The solution is completely analogous. | No |
Problem 5.6.13. Let \( {ABCD} \) be a cyclic quadrilateral with circumcenter \( O \) . The points \( {O}_{1},{O}_{2},{O}_{3} \) and \( {O}_{4} \) are the circumcenters of \( \bigtriangleup {ABO} \) , \( \bigtriangleup {BCO},\bigtriangleup {CDO} \) and \( \bigtriangleup {DAO} \), respectively. Let \( {AC} \cap {BD} = P ... | Solution. Let the circumcircles of \( \bigtriangleup {ABO} \) and \( \bigtriangleup {CDO} \) intersect at the points \( O \) and \( F \), and let the circumcircles of \( \bigtriangleup {ADO} \) and \( \bigtriangleup {BCO} \) intersect at the points \( O \) and \( E \) .\n\nProblem 5.6.4 yields that \( \angle {PEO} = \a... | Yes |
Problem 5.6.14. Let \( {ABCD} \) be a cyclic quadrilateral with circumcenter \( O \) . Let \( {AC} \cap {BD} = P \) . The points \( {O}_{1},{O}_{2},{O}_{3} \) and \( {O}_{4} \) are the cir-cumcenters of \( \bigtriangleup {ABP},\bigtriangleup {BCP},\bigtriangleup {CDP} \) and \( \bigtriangleup {DAP} \), respectively. Pr... | Solution. Let the circumcircles of \( \bigtriangleup {ABP} \) and \( \bigtriangleup {CDP} \) intersect at the points \( P \) and \( F \), and let the circumcircles of \( \bigtriangleup {ADP} \) and \( \bigtriangleup {BCP} \) intersect at the points \( P \) and \( E \) .\n\nProblem 5.6.2 yields that \( \angle {PEO} = \a... | Yes |
Problem 5.6.15. Let \( {ABCD} \) be a cyclic quadrilateral with circumcenter \( O \) . Let \( {AC} \cap {BD} = P \) . The points \( {O}_{1} \) and \( {O}_{2} \) are the circum-centers of \( \bigtriangleup {ADP} \) and \( \bigtriangleup {BCP} \), respectively. Prove that the quadrilateral \( {\mathrm{O}}_{1}{\mathrm{{OO... | Solution. Note that \( O{O}_{1} \bot {AD} \), since \( O{O}_{1} \) is the perpendicular bisec- tor of \( {AD} \) . Moreover, we have \[ \angle {O}_{2}{PB} = {90}^{ \circ } - \angle {PCB} = {90}^{ \circ } - \angle {ADP} \Rightarrow {O}_{2}P \bot {AD} \Rightarrow {O}_{2}P\parallel O{O}_{1}\text{.} \] Analogously, we dedu... | Yes |
Problem 5.6.16. Let \( {ADBC} \) be a cyclic quadrilateral with circumcenter \( O \) . Let \( {AC} \cap {BD} = P \) . The points \( {O}_{1} \) and \( {O}_{2} \) are the circum-centers of \( \bigtriangleup {ADP} \) and \( \bigtriangleup {BCP} \), respectively. Prove that the quadrilateral \( {\mathrm{O}}_{1}{\mathrm{{OO... | Solution. This problem is essentially the same as Problem 5.6.15 but for a quadrilateral \( {ADBC} \) instead of \( {ABCD} \) . The solution is completely analogous. | No |
Problem 5.6.17. Let \( {ABCD} \) be a cyclic quadrilateral with circumcenter \( O \) . Let \( {AC} \cap {BD} = P \) . The circumcircles of \( \bigtriangleup {ADO} \) and \( \bigtriangleup {BCO} \) intersect for the second time at the point \( Q \) . Prove that the quadrilaterals \( {QPCD} \) and \( {ABPQ} \) are cyclic... | Solution. We have\n\n\[ \angle {CQD} = {360}_{.}^{ \circ } - \angle {CQO} - \angle {DQO} = \angle {OBC} + \angle {OAD} \]\n\n\[ = {90}^{ \circ } - \angle {BAC} + {90}^{ \circ } - \angle {DBA} = {180}^{ \circ } - \angle {PAB} - \angle {PBA} \]\n\n\[ = \angle {APB} = \angle {CPD}\text{. } \]\n\nThus, the quadrilateral \(... | Yes |
Problem 5.6.18. Let \( {ABDC} \) be a cyclic quadrilateral with circumcenter \( O \) . Let \( {AC} \cap {BD} = P \) . The circumcircles of \( \bigtriangleup {ADO} \) and \( \bigtriangleup {BCO} \) intersect for the second time at the point \( Q \) . Prove that the quadrilaterals \( {QPCD} \) and \( {ABPQ} \) are cyclic... | Solution. This problem is essentially the same as Problem 5.6.17 but for a quadrilateral \( {ABDC} \) instead of \( {ABCD} \) . The solution is completely analogous. | No |
Problem 5.7.1. Let \( {ABCD} \) be a quadrilateral with \( {AC} \cap {BD} = E \) . Let \( M \) and \( Q \) be the projections of the points \( A \) and \( D \) onto the line \( {BC} \) , respectively. Let \( N \) and \( P \) be the projections of the points \( B \) and \( C \) onto the line \( {AC} \), respectively. If... | Solution. We have that the quadrilaterals \( {AMCP} \) and \( {BNDQ} \) are inscribed in the circles \( {k}_{1} \) and \( {k}_{2} \), respectively. On the other hand, \( \bigtriangleup {DPY} \sim \bigtriangleup {CQY} \), which gives \( {DY}.{YQ} = {PY}.{YC} \) . Also, \( \bigtriangleup {AXN} \sim \bigtriangleup {BXM} \... | Yes |
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