In the Quadrilateral $ABCD$ we have $ \measuredangle B=\measuredangle D = 60^\circ $.$M$ is midpoint of side $AD$.The line through $M$ parallel to $CD$ meets $BC$ at $P$.Point $X$ lying on $CD$ such that $BX=MX$.Prove that $AB=BP$ if and only if $\measuredangle MXB=60^\circ$. Author: Davoud Vakili, Iran
Problem
Source: Iranian geometry olympiad 2014
Tags: geometry, geometry proposed
TelvCohl
22.02.2015 21:25
See quadrilateral -prove
Strate
05.10.2022 18:27
Suppose that $AB=BP$, this implies that $ABP$ is an equilateral triangle since $\measuredangle B = 60$°. Define a rotation centered at $B$ with angle $60$° that sends $A$ to $P$. We claim that it sends $M$ to $X$. To show this, call $X'$ the image of $M$, and $C'$, $D'$ the intersection of the parallel through $X'$ of $(CD)$ and $(BP), (AM)$ respectively. To show that $X=X'$, it is enough to show that $D=D'$ which can be done by showing that $M$ is the midpoint of $\overline{AD'}$, i.e that $MD'=MA=PX'$. So we want to show that $MPX'D'$ cyclic. Let $P'$ the image of $P$ by the rotation, and $T=(C'D') \cap (BP')$. Since $\angle AD'T=60$ and $\angle ABT= 120$ it follows that $ABTD'$ is cyclic. But $ \measuredangle BTX' = \measuredangle BTD' = \measuredangle BAD'=\measuredangle BAM = \measuredangle BPX'$ since rotation preserve angles. Hence $BPX'T$ is cyclic and it follows that $\angle PX'D'=60$ and that $PMD'X'$ is cyclic. Hence $D=D'$, $C=C'$, $X=X'$ and we are done. The reciprocal can be shown in an analogous manner.
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Tsikaloudakis
15.10.2022 09:38
στα υποψιν
Om245
03.10.2024 11:00
Fantastic If $AB=BP$ Note $\triangle ABP$ is equilateral triangle and $D,A,P,C$ cyclic. Now let $K = AP \cap CD$. Note $MB$ is symmedian in $\triangle AMP$ and $\triangle AMP \sim \triangle ADK$ in which $DP$ is median., hence $$\angle BMA = \angle CDP = \angle MPD$$Consider point $X$ on $CD$ such that $M,P,X,D$ cyclic. Now as $MP \parallel DX$ we get $AM=MD=PX$ also $\angle PXD = \angle AMP = 60$. Hence $\angle BPX = \angle PXD + \angle PCX = \angle BAD + 60$. Observe as $PX = AM$, $\angle BPX = \angle BAM$ and $AB = BP$, hence $\triangle BAM \cong \triangle BPX$ which give us $BP = BX$ and $\angle MBX = 60$. So we get $\triangle MXB$ is equilateral triangle. If $\measuredangle MXB = 60$. Now we redefine $P$ here. Define $X$ as given. Let $P$ be point on $(MXD)$ such that $MP \parallel XD$. Note now we have using $\angle AMP = 60$ that $$\angle AMB = \angle XMP = \angle XDP = \angle BPD = \angle PXB$$Also $AM=MD=PX$. As $BM = BX$ we get $\triangle MAB \cong \triangle XPB$ which give us $AB = BP$ and $\angle ABP = 60$. hence $P$ is indeed problems point and we have $BA=BP$. Here is moment while solving this on Microsoft Paint
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