prove a triangle is isoceles

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prove a triangle is isoceles

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Source: Czech and Slovak third round,2007,p2

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#1 h Mar 3, 2012, 5:06 AM • 2 Y

Y by Adventure10, Mango247

In a cyclic quadrilateral $ABCD$ , let $L$ and $M$ be the incenters of $ABC$ and $BCD$ respectively. Let $R$ be a point on the plane such that $LR\bot AC$ and $MR\bot BD$ .Prove that triangle $LMR$ is isosceles.

This post has been edited 1 time. Last edited by Amir Hossein, Mar 3, 2012, 6:35 AM
Reason: Fixed LaTeX.

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#2 h Mar 3, 2012, 7:07 AM • 2 Y

Y by Adventure10, Mango247

We prove that $RM =RL$ , equivalently, $\angle RML = \angle RLM$

Let $X$ and $Y$ be the intersections of $ML$ with $AC$ and $BD$ respectively.

So the problem is true iff $\angle DYM =\angle AXL \leftrightarrow \angle LYB = \angle MXC$

The last equality is equivalent to $\angle LMC + \angle MCA = \angle MLB + \angle DBL$ (*).

Since $\angle BDC = \angle BAC$ then $\angle BLC = \angle BMC = 90 + \frac{\angle BAC}{2}$ so $BLMC$ is cyclic.

So $\angle LMC + \angle MCA = 180 - \angle LBC + \angle DCA - \angle DCM$

Also $\angle MLB + \angle DBL = 180 -\angle MCB + \angle DBA - \angle ABL$

So equality in (*) is equivalent to

$180 - \angle LBC + \angle DCA - \angle DCM = 180 - \angle MCB + \angle DBA -\angle ABL$

Which is true because $\angle LBC = \angle ABL$ , $\angle DCM = \angle MCB$ and $\angle DCA = \angle DBA$ and we are done.

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