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Prove that the lengths of tangents drawn from an external point $A$ to the points $P$ and $Q$ on the circle are equal.

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Answer:

Step by Step Explanation:

1. It is given that two tangents are drawn from an external point $A$ to the points $P$ and $Q$ on the circle.
   
   The given situation is represented by the below image.
   
   
   
   We have to prove that the length $AP$ is equal to length $AQ$.
2. Let us join the point $O$ to points $P, Q,$ and $A.$
   We get
   
   $AP$ is a tangent at $P$ and $OP$ is the radius through $P$.
   We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
   $\implies OP \perp AP$
   
   Also, $AQ$ is a tangent at $Q$ and $OQ$ is the radius through $Q$.
   We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
   $\implies OQ \perp AQ$
3. In right- angled triangle $OPA$ and $OQA$, we have $$\begin{aligned} & OP = OQ && \text{[Radius of the same circle]} \\ & OA = OA && \text{[Common]} \\ \implies & \triangle OPA \cong \triangle OQA && \text{[By RHS-congruence]} \end{aligned}$$
4. As the corresponding parts of congruent triangle are equal, we have $AP = AQ$.
5. Thus, the lengths of tangents drawn from an external point $A$ to the points $P$ and $Q$ on the circle are equal.

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