Chapter 9.2, Problem 15WE

Solution

To prove: $PR+RS+SP=PA+PB$

Given:

The figure:

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

Concept Used:

Theorem 9-1 Corollary: Tangents to a circle from a point are congruent.

Explanation:

Consider the given figure:

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 (1) and (2),

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

Hence proved