Question

Given: $\overline{RS}$ is a common internal tangent to $\odot A$ and $\odot B$.

Explain why $\frac{AC}{BC}=\frac{RC}{SC}$.

Short answer

$\frac{AC}{BC}=\frac{RC}{SC}$

Step-by-step solution

Step 1. Given information.

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

Step 2. Concept used.

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,

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

Then,

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

Step 3. First prove that the two triangles are similar.

Consider in $\Delta ARC$and $\Delta BSC$

$\angle ARC=\angle BSC$(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, $\frac{AC}{BC}=\frac{RC}{SC}$