Two circles are externally tangent to each other at a point $A$ and internally tangent to a third circle $\Gamma$ at points $B$ and $C.$ Let $D$ be the midpoint of the secant of $\Gamma$ which is tangent to the smaller circles at $A.$ Show that $A$ is the incenter of the triangle $BCD$ if the centers of the circles are not collinear.
Problem
Source: Turkey National Olympiad 2002 - D1 - P2
Tags: geometry, incenter, circumcircle, angle bisector, geometry unsolved
mousavi
11.03.2011 16:00
let $O$ be the circumcenter of the biggest circle.we want to prove that the quadrilateral $BODC$ is cyclic.draw the tangents at points $B,C$.thse two tangents and tangents at point $A$ are concurrent at point $P$.$\angle PDO=\angle OBP=90$, so $BPDO$ is cyclic.and $\angle ODP=\angle OCP=90$, so $OPCD$ is cyclic.$\angle OCD=\angle DPO=\angle DOB$ so $BODC$ is cyclic.then,$\angle PDB=\angle POB$ and $\angle PDC=\angle POC$ so $\angle CDP=\angle BDP$.points $B,A,C$ are on a circle with center $P$ so $\angle BPA=2\angle BCA$ and $\angle BPA=\angle BCD$ so $CA$ is angle bisector of $\angle BCD.$
mrdriller
18.04.2021 15:47
Let $ AD $ meets $\Gamma $ at $ P $ and $ Q $. Let $ BA $ intersects $\Gamma $ again $ E $ and $ CA $ intersects $\Gamma $ again at $ F $. Lemma $ E $ is the midpoint of the arc $ PCQ $. Proof There is a homothety centered $ B $ and takes $\omega_1 $ to $\Gamma $ and $ A $ to $ E $. Then $ E $ is clearly the midpoint of the arc $ PCQ $. So $ E $ and $ F $ are the midpoints of the arcs $ PCQ $ and $ PBQ $ respectively. So $ EF $ is the diameter of the circle which passes through $ D $. So $ \angle FBE= \angle FCE= 90^{\circ} $. Since $\angle PDE= \angle ADE= 90^{\circ} $, $ ADEC $ is a cyclic quadrilateral. We got everything we need. $$\angle BCA=\angle BCF= \angle BEF= \angle AED= \angle ACD $$So $ CA $ is the angle bisector of $\angle BCD $. We can prove similarly prove $ \angle CBA= \angle ABD $. So A is the incenter of $ BCD $. $\blacksquare$