Question

Given $\overline{PA}$and $\overline{PB}$ are tangents to $\odot o.$

Use the diagram at the right to explain how the corollary on page $333$

follows from Theorem$9-1.$

Short answer

a. Tangent $PA$ is congruent to tangent $PB.$

Step-by-step solution

Step 1. Given information:

The figure is given.

Step 2. Concept used:

We used the theorem 9-1.

Step 3. Applying the concept:

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

Then,

$\begin{array}{l}OA\perp PA\\ OB\perp PB\end{array}$

Then once can say that,

$\begin{array}{l}\angle PAO={90}^{\circ }\\ \angle PBO={90}^{\circ }\end{array}$
It is required to prove two triangles are congruent.

Consider in $▵PAO$ and $▵PBO,$

$\begin{array}{l}OA=OB\\ \angle PAO=\angle PBO\\ PO=OP\end{array}$

(Radius of a circle is equal)

(Both the angles are )

(Common side of a triangle)

So, it implies that

$▵PAO\cong ▵PBO$ ( Side-Angle-Side Rule)

Since the corresponding parts of congruent triangles are congruent.

Therefore, the following holds good $PA$ and $PB.$