Let $k_1$ and $k_2$ be two circles and let them cut each other at points $A$ and $B$. A line through $B$ is cutting $k_1$ and $k_2$ in $C$ and $D$ respectively, such that $C$ doesn't lie inside of $k_2$ and $D$ doesn't lie inside of $k_1$. Let $M$ be the intersection point of the tangent lines to $k_1$ and $k_2$ that are passing through $C$ and $D$, respectively. Let $P$ be the intersection of the lines $AM$ and $CD$. The tangent line to $k_1$ passing through $B$ intersects $AD$ in point $L$. The tangent line to $k_2$ passing through $B$ intersects $AC$ in point $K$. Let $KP \cap MD \equiv N$ and $LP \cap MC \equiv Q$. Prove that $MNPQ$ is a parallelogram.
Problem
Source: Macedonia National Olympiad 2015
Tags: geometry
TelvCohl
05.04.2015 20:40
My solution: Since $ \angle MCD=\angle CAB, \angle MDC=\angle DAB $ , so $ \angle CAD+\angle DMC=180^{\circ} \Longrightarrow A, C, D, M $ are concyclic , hence combine with $ \angle MAD=\angle MCD=\angle CAB $ we get $ \triangle ACB \sim \triangle AMD $ , so $ \angle PAL=\angle CAB=\angle DBA-\angle DCA=\angle DBA-\angle LBA=\angle PBL \Longrightarrow L \in \odot (ABP) $ . Similarly we can prove $ K $ lie on $ \odot (ABP) \Longrightarrow A, B, K, L, P $ are concyclic . From Reim theorem ( for $ A-L-D $ and $ B-P-D $ ) we get $ LP \parallel DM $ . Similarly we can prove $ KP \parallel CM \Longrightarrow MNPQ $ is a parallelogram . Q.E.D
andria
05.04.2015 21:48
My solution: $\angle ADM=\angle ADC+\angle CDM=\angle ADC+\angle BAD=\angle ABC=180-\angle ACM$ so quadrilateral $ ACMD $ is cyclic also $180=\angle DCM+\angle LBC=\angle DAM+\angle LBC=180$ so quadrilateral $ APBL $ is cyclic too note that $\angle AMD=\angle ACD=\angle ABL=\angle APL \longrightarrow QL||MD $ and $\angle CAM=\angle CDM=\angle KBC $ so $ AKPB $ is cyclic $\longrightarrow \angle KPA=\angle KBA=\angle ADC=\angle AMC \longrightarrow KN||CM $ DONE