Tangent circles and isosceles triangle

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anantmudgal09

1980 posts

anantmudgal09

#1 h Apr 28, 2018, 7:58 PM • 3 Y

Y by g0USinsane777, Adventure10, Mango247

Circle $\omega$ is tangent to sides $AB, AC$ of triangle $ABC$ . A circle $\Omega$ touches the side $AC$ and line $AB$ (produced beyond $B$ ), and touches $\omega$ at a point $L$ on side $BC$ . Line $AL$ meets $\omega, \Omega$ again at $K, M$ . It turned out that $KB \parallel CM$ . Prove that $\triangle LCM$ is isosceles.

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WolfusA

1900 posts

WolfusA

#2 h Apr 28, 2018, 8:12 PM • 1 Y

Y by Adventure10

Shouldn't it be "touches $\omega$ at a point $L$ on side $AC$ ?
I don't get the configuration at all.

This post has been edited 1 time. Last edited by WolfusA, Apr 28, 2018, 8:14 PM

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falantrng

250 posts

falantrng

#3 h Apr 28, 2018, 8:14 PM • 3 Y

Y by Abbas11235, Kristal-jump, Adventure10

WolfusA wrote:

Shouldn't it be "touches $\omega$ at a point $L$ on side $AC$ ?

But $AL$ cannot intersects $\omega,\Omega.$
Problem condition is wrong.Please be careful.

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anantmudgal09

1980 posts

anantmudgal09

#4 h Apr 28, 2018, 8:16 PM • 4 Y

Y by hellomath010118, third_one_is_jerk, Adventure10, Mango247

Note that dilation at $L$ maps $\omega \mapsto \Omega$ hence $K \mapsto M$ and $B \mapsto C$ . Thus the line $CA'$ through $C$ parallel to $AB$ is a tangent to $\Omega$ . Let $O$ be the center of $\Omega$ then line $CO$ bisects angle $A'CA$ . Meanwhile line $AO$ bisects angle $BAC$ hence $\angle AOC=90^{\circ}$ . Since $ML$ is a diameter of $\Omega$ and $OC \perp ML$ we conclude $CL=CM$ .

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Mprog.

39 posts

Mprog.

#6 h Aug 15, 2018, 3:27 PM • 2 Y

Y by Adventure10, Mango247

anantmudgal09 wrote:

Note that dilation at $L$ maps $\omega \mapsto \Omega$ hence $K \mapsto M$ and $B \mapsto C$ . Thus the line $CA'$ through $C$ parallel to $AB$ is a tangent to $\Omega$ . Let $O$ be the center of $\Omega$ then line $CO$ bisects angle $A'CA$ . Meanwhile line $AO$ bisects angle $BAC$ hence $\angle AOC=90^{\circ}$ . Since $ML$ is a diameter of $\Omega$ and $OC \perp ML$ we conclude $CL=CM$ .

Can someone explain this solution in detail, please, or show another solution?

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