Let $a$ and $b$ be two circles, intersecting in two distinct points $Y$ and $Z$. A circle $k$ touches the circles $a$ and $b$ externally in the points $A$ and $B$. Show that the angular bisectors of the angles $\angle ZAY$ and $\angle YBZ$ intersect on the line $YZ$.
Problem
Source: Germany 2019, Problem 2
Tags: geometry
Pluto1708
20.06.2019 13:40
Easy
Let $T=AA\cap BB$.Note that it suffices to show $\dfrac{AY}{AZ}=\dfrac{BY}{BZ}$.By radical axis its easy to see that $T\in YZ$.Since $\dfrac{AY}{AZ}=\dfrac{TY}{TA}=\dfrac{TY}{TB}=\dfrac{BY}{BZ}$.So we are done.$\blacksquare$
Mogmog8
27.05.2022 08:28
Same as @above; posting for storage. By Radical axis, let $X=\overline{AA}\cap\overline{BB}\cap\overline{YZ}.$ Then, $\triangle AXZ\sim\triangle YXA$ and $\triangle BXZ\sim\triangle YXB$ so $$\frac{AY}{AZ}=\frac{AX}{XZ}=\frac{BX}{XZ}=\frac{BY}{BZ}.$$The Angle Bisector Theorem finishes. $\square$
SouradipClash_03
12.12.2023 12:06
Perform an inversion on $Y$ with any radius. Circles $a$ and $b$ turn to straight lines $a'$ and $b'$, that intersect on a point $Z'$. Note that $Z'$ lies on $YZ$. Circle $k$ turn to a circle $k'$ such that its tangent to the lines $a'$ and $b'$, at the points $A'$ and $B'$. By properties of tangency, $$\angle A'Z'Y = \angle B'Z'Y$$$$\implies \angle AZY = \angle BZY$$Thus $YZ$ is the angle bisector, and thus the other two angle bisectors will meet at the incenter $I$ of $\triangle AZB$, which lies on the angle bisector $YZ$.