Let $ABC$ be a triangle. $M$, and $N$ points on $BC$, such that $BM=CN$, with $M$ in the interior of $BN$. Let $P$ and $Q$ be points in $AN$ and $AM$ respectively such that $\angle PMC= \angle MAB$, and $\angle QNB= \angle NAC$. Prove that $ \angle QBC= \angle PCB$.
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Tags: geometry, national olympiad, geometry unsolved, geometry proposed
TelvCohl
28.04.2015 08:51
My solution: Let $ A^*, P^* $ be the reflection of $ A, P $ in the perpendicular bisector of $ BC $, respectively . Let $ X=NQ \cap A^*B $ and $ Y=NP^* \cap AB $ (Easy to see $ P^* \in A^*M $ (from symmetry)) . Since $ AA^*MN $ is an Isosceles trapezoid , so $ A, A^*, M, N $ are concyclic . ... $ (1) $ Since $ \angle XNM=\angle NAC=\angle XA^*M $ , so $ A^*, X, M, N $ are concyclic . ... $ (2) $ Since $ \angle YNM=\angle NMP=\angle YAM $ , so $ A, Y, N, M $ are concyclic . ... $ (3) $ From $ (1), (2), (3) \Longrightarrow A, A^*, M, N, X, Y $ are concyclic , so from Pascal theorem ( for $ AYNXA^*M $ ) we get $ P^* \in BQ \Longrightarrow \angle QBC=\angle P^*BC=\angle PCB $ . Q.E.D
ShinyDitto
06.02.2017 03:45
Assume WLOG $AC > AB$. Let $Z \equiv MP \cap AC$. Let $X$ be a point on the ray $QN$ with $Q$ between $X$ and $N$ such that $\measuredangle NAX = \measuredangle MAC$. $\Longrightarrow \measuredangle ACB= \measuredangle ANQ$. $\Longrightarrow \triangle ACM \sim \triangle ANX \Longrightarrow \tfrac{AC}{AN} = \tfrac{MC}{XN} = \tfrac{NB}{XN} \Longrightarrow \triangle NAC \sim \triangle XNB$. Furthermore $\measuredangle XBN = \measuredangle NCA$. $$\Longrightarrow \frac{NC}{AC} = \frac{BX}{BN}$$$$\Longrightarrow \frac{AB}{AC} \cdot \frac{BM}{AB} = \frac{BX}{BN}$$$$\Longrightarrow \frac{ \sin \measuredangle ACB}{\sin \measuredangle ABC} \cdot \frac{ \sin \measuredangle BAM}{ \sin \measuredangle AMC} = \frac{BX}{BN}$$$$\Longrightarrow \frac{ \sin \measuredangle ACB}{ \sin \measuredangle AMZ} \cdot \frac{ \sin \measuredangle ZMC}{ \sin \measuredangle AMC} = \frac{BX}{BN}$$$$\Longrightarrow \frac{ \sin \measuredangle (AC, MZ)}{ \sin \measuredangle AMZ} \cdot \frac{ \sin \measuredangle ZMC}{\sin \measuredangle (AC, MZ)} = \frac{ \sin \measuredangle AMC}{ \sin\measuredangle ACB} \cdot \frac{BX}{BN}$$$$\Longrightarrow \frac{AM}{AZ} \cdot \frac{CZ}{CM} = \frac{AC}{AM} \cdot \frac{BX}{BN}$$$$\Longrightarrow \frac{MP}{PZ} \cdot \frac{ \sin \measuredangle CAN}{ \sin\measuredangle NAM} \cdot \frac{CZ}{CM} = \frac{AN}{AX} \cdot \frac{BX}{BN}$$$$\Longrightarrow \frac{MP}{PZ} \cdot \frac{ \sin \measuredangle CAN}{ \sin \measuredangle NAM} \cdot \frac{CZ}{CM} = \frac{NQ}{XQ} \cdot \frac{ \sin \measuredangle QAX}{ \sin \measuredangle NAM} \cdot \frac{BX}{BN}$$$$\Longrightarrow \frac{MP}{PZ} \cdot \frac{CZ}{CM} = \frac{NQ}{XQ} \cdot \frac{BX}{BN}$$$$\Longrightarrow \frac{ \sin \measuredangle MCP}{ \sin \measuredangle ZCP} = \frac{ \sin \measuredangle NBQ}{ \sin \measuredangle XBQ}$$$\measuredangle MCP + \measuredangle ZCP = \measuredangle NBQ+ \measuredangle XBQ$ By Two Equal Angles Lemma, it follow $\measuredangle MCP = \measuredangle NBQ$