Prove that the lengths of tangents drawn from an external point ^@ A ^@ to the points ^@ P ^@ and ^@ Q ^@ on the circle are equal.
Answer:
- 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 ^@. - 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 ^@ - 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} @^
- As the corresponding parts of congruent triangle are equal, we have ^@ AP = AQ ^@.
- Thus, the lengths of tangents drawn from an external point ^@ A ^@ to the points ^@ P ^@ and ^@ Q ^@ on the circle are equal.