Q15. - Geometry

Question

$\overline{PA}$, $\overline{PB},$ and $\overline{RS}$ are tangents.

Explain, Why $PR+RS+SP=PA+PB.$

Step-by-Step Solution

Answer

$PR+RS+SP=PA+PB$

Step 1. Given information.

The figure here given is,

$\overline{PA}$, $\overline{PB}$and $\overline{RS}$are tangents 

Step 2. Concept Used.

Theorem $9.1$ Corollary,

 Tangents to a circle from a point are congruent.

Step 3. Consider the given figure for further solution.

From the above figure,

 it is clear that, $\overline{RA}$ and $\overline{RC}$ are tangents to the circle from the common point R, so by theorem $9.1$ corollary, it can be said that,

$RA=RC$                                                                                   ….. (1)

Similarly, $\overline{BS}$ and $\overline{CS}$ are tangents to the circle from the common point \[S\], so by theorem $9.1$ corollary, it can be said that,

$BS=CS$                                                                                   ….. (2)

Now consider,

$PA+PB=PR+RA+BS+SP$

Then using $\left(1\right)$ and $\left(2\right)$,

$\begin{array}{c}PA+PB=PR+RC+CS+SP\\ =PR+\left(RC+CS\right)+SP\\ =PR+RS+SP\end{array}$

Therefore, it is proved that, $PR+RS+SP=PA+PB$.