angle bisectors in trapezoid

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angle bisectors in trapezoid

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Source: Ukrainian TST 2007 problem 2

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#1 h Sep 19, 2007, 12:51 PM • 2 Y

Y by Adventure10, Mango247

$ABCD$ is convex $AD\parallel BC$ , $AC\perp BD$ . $M$ is interior point of $ABCD$ which is not a intersection of diagonals $AC$ and $BD$ such that $\angle AMB =\angle CMD =\frac{\pi}{2}$ . $P$ is intersection of angel bisectors of $\angle A$ and $\angle C$ . $Q$ is intersection of angel bisectors of $\angle B$ and $\angle D$ . Prove that $\angle PMB =\angle QMC$ .

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#2 h Sep 27, 2007, 12:46 PM • 2 Y

Y by Adventure10, Mango247

Denote intersection point of diogonals,intersectıon poınt of angle bısectors $\angle B$ and $\angle A$ and intersection point of angle bısectors $\angle C$ and $\angle D$ by $L$ , $Y$ and $X$ ,respectıvely.
$AD\parallel BC\Rightarrow\angle CXD=\angle AYB=\frac{\pi}{2}$ .
So five points $B,L,Y,M$ and $A$ are cyclıc. As the same $C,L,X,M$ and $D$ are also cyclıc.Then
$\angle XMY=\angle XML+\angle YML=\angle XDL+\angle YAL=\angle XCL+\angle YAL=\angle XPY$
$\Rightarrow X,Y,Q,M,P$ is cyclıc.
It ıs easy to show that $XY\parallel AD$ .So
$\angle YPQ=\angle YXQ=\angle QDA\Rightarrow PQDA$ is cyclic.
From all of these
$\angle PMB=\angle PMY-\angle BMY$ $(1)$
$\angle QMC=\angle QMX-\angle CMX$ $(2)$
$\Rightarrow\angle PMB-\angle QMC=\angle PMY-\angle QMX+\angle CMX-\angle BMY=\angle PMX-\angle QMY+\angle XDC-\angle YAB=\angle PQX-\angle YPQ+\angle XDA-\angle YAD=0$
$\Rightarrow\angle PMB=\angle QMC$

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