Question

Let $\Omega$ be a circle with a chord $AB$ which is not a diameter. Let ${\Gamma }_1$ be a circle on one side of $AB$ such that it is tangent to $AB$ at $C$ and internally tangent to $\Omega \text{at}D$. Likewise, let ${\Gamma }_2$ be a circle on the other side of $AB$ such that it is tangent to $AB$ at $E$ and internally tangent to $\Omega \text{at}F$. Suppose the line $DC$ intersects $\Omega \text{at}X\ne D$ and the line $FE$ intersects $\Omega atY\ne F$. Prove that $XY$ is a diameter of $\Omega$.

Answer 1

Join $DY$ and $FX$

Join tangent from $D$ with line $AB$ by extending it and join tangent at $F$ with line $AB$ by extending

Let $\angle PDX=\alpha \angle P_{1}DY=\beta \angle QFY=\gamma and\angle Q_{1}FX=\delta$

Using alternate theorem of segment in circle $\Omega$ we get

$\angle DXY=\angle P_{1}DY=\beta$ and $\angle FYX=\angle Q_{1}FX=\delta$

$PD$ and $PC$ are tangents from $P$ to the ${\Gamma }_1$

$PD=PC$

$\angle PCD=\angle PDC=\angle PDX=\alpha$

$\angle XCR=\angle PCD=\alpha$

In $△XC{R}_1$

$\angle XRC$$={180}^o-(\angle CXR+\angle XCR)={180}^o-(\alpha +\beta )\left(1\right)$

Similarly, in $△ERY$

$\angle RYE=\angle FYX=\delta$ and $\angle REY=\angle FEQ=\angle QFE=\gamma$

$\angle ERY={180}^o-(\gamma +\delta )\left(2\right)$

but $\angle XRC=\angle ERY$

From $\left(1\right)$ and $\left(2\right)$

$\alpha +\beta =\gamma +\delta \left(3\right)$

Now, $DXFY$ is a cyclic quadrilateral

$\angle XDY+\angle XFY={180}^o{180}^o-(\alpha +\beta )+{180}^o-(\gamma +\delta )={180}^o\alpha +\beta +\gamma +\delta ={180}^o$

From $\left(3\right)$

$2(\alpha +\beta )={180}^o\Rightarrow \alpha +\beta ={90}^o$

$\angle XDY={180}^o-(\alpha +\beta )={90}^o$ (angle in semicircle)

Hence, $XY$ is a diameter.