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.
    O A P Q


    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
    O A P Q
    ^@ 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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