Chapter 9.2, Problem 11WE

To determine

To Prove: The ratio of the $\frac{AC}{BC}=\frac{RC}{SC}$ for the given figure .

Explanation of Solution

Given information:

  

  $\overline{RS}$ is a common internal tangent to the circles with center A and B.

If the line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of tangents.

This implies that,

  $\begin{array}{l}AR\perp RS\hfill \\ SB\perp RS\hfill \end{array}$

Then,

  $\begin{array}{l}\angle ARC=90°\hfill \\ \angle BSC=90°\hfill \end{array}$

First prove that the two triangles are similar.

Consider in ?ARC and ?BSC

  $\angle ARC=\angle BSC$ (Both the angles are of measure 90°)

  $\angle ACR=\angle BCR$ (Vertical opposite angles is equal)

This implies that, $\Delta ARC~\Delta BSC$ (Angle-Angle Rule)

The corresponding sides in similar triangle are proportional.

Therefore, it is proved that $\frac{AC}{BC}=\frac{RC}{SC}$