In the convex quadrilateral $ABCD$ we have that $\angle BCD = \angle ADC \ge 90 ^o$. The bisectors of $\angle BAD$ and $\angle ABC$ intersect in $M$. Prove that if $M \in CD$, then $M$ is the middle of $CD$.
Problem
Source: 2011 Romanian NMO grade VII P3
Tags: geometry, angle bisector, midpoint, equal angles
Gryphos
18.05.2020 09:23
Since $AD \perp CD \perp BC$, we have $$CM = \operatorname{dist}(M,BC)= \operatorname{dist}(M,AB)= \operatorname{dist}(M,AD)=DM.$$Edited: Misread question.
ak_47
18.05.2020 09:42
Gryphos wrote: Since $AD \perp CD \perp BC$, we have $$CM = \operatorname{dist}(M,BC)= \operatorname{dist}(M,AB)= \operatorname{dist}(M,AD)=DM.$$ How did you get $AD \perp CD \perp BC$ The given condition is $\angle BCD = \angle ADC \ge 90 ^o$
ak_47
18.05.2020 09:50
Let $AD$ and $BC$ extended meet at $K$
Hence, $M$ is the incenter of $\triangle ABK$
Let $MY \perp AK$ and $MX \perp BK$ with $Y, X$ being on $AK, BK$ respectively.
As $M$ is incenter, $MY=MX, KX=KY$
As $\angle BCD = \angle ADC \implies \angle CDK = \angle DCK$
Hence, $\angle MDY=\angle MCX, MY=MX, \angle MYD = \angle MXC = 90$
Hence, $\triangle MDY \cong \triangle MCX \implies MD=MC$