Question

Two circles with centres X and Y touch each other externally at P. Two diameters AB and CD are drawn one in each circle parallel to other. Prove that B, P and C are collinear.

Answer 1

We know,

$\angle APB={90}^o,\angle CPD={90}^o$ (angle inscribed in a semicircle)

We know $XPY$ is a straight line

In $△XPB,XP=XB$ (radii)

$\angle XPB=\angle XBP$ (angle opposite to equal sides)

$\angle XPB=\angle CPY$ (opposite angle) $\left(1\right)$

In $△CPYPY=CY$ (radii)

$\angle YCP=\angle YPC$ (angle opposite to equal sides) $\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get

$\angle XBP=\angle YCP\left(3\right)$

$\Rightarrow CY\parallel XB$ as ($\angle XBP,\angle YCP$ are a;ternate angles)

$\frac{XP}{XB}=\frac{YC}{YP}=1\left(4\right)$

From $\left(3\right)$ and $\left(4\right)$ we can conclude that

$B,P,C$ are collinear