From a point ^@ P ^@, two tangents ^@ PA ^@ and ^@ PB ^@ are drawn to a circle ^@ C(O,r) ^@. If ^@ OP = 2r ^@, show that ^@ \triangle APB ^@ is an equilateral triangle.


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Answer:


Step by Step Explanation:
  1. Let ^@ OP ^@ meet the circle at ^@ Q ^@. Join ^@ OA ^@ and ^@ AQ ^@.
    A O Q P B
  2. We know that the radius through the point of contact is perpendicular to the tangent. @^ \begin{aligned} \text{ So, } OA \perp AP \implies \angle OAP = 90^\circ && \ldots \text{(i)} \end{aligned} @^
  3. The circle is represented as ^@ C(O,r) ^@, this means that ^@ O ^@ is the center of the circle and ^@ r ^@ is its radius. @^ \begin{aligned} \implies OQ = OA = r && \ldots \text{(ii)} \end{aligned} @^ Also, we see that ^@ OP = OQ + QP. ^@

    Substituting the value of ^@ OP ^@ and ^@ OQ ^@ in the above equation, we have @^ \begin{aligned} & OP = OQ + QP \\ \implies & 2r = r + QP \\ \implies & QP = 2r - r = r \\ \implies & \text{ Q is the mid-point of OP.} && \text{ [As } QP = OQ = r ] \end{aligned} @^
  4. As, ^@ Q ^@ is the mid-point of ^@ OP, AQ ^@ is the median from the vertex ^@ A ^@ to the hypotenuse ^@ OP ^@ of the right-angled triangle ^@ AOQ ^@.

    We know that the median on the hypotenuse of a right- angled triangle is half of its hypotenuse.
    Thus, @^ QA = \dfrac { 1 } { 2 } OP = \dfrac { 1 } { 2 } (2r) = r. @^ ^@ \begin{aligned} \implies & QA = OQ = QP = r \\ \implies & OA = OQ = QA = r && \text{[Using } eq \text{ (ii)}] \\ \implies & \triangle AOQ \text{ is an equilateral triangle. } \\ \implies & \angle AOQ = 60^\circ && \text{ [Each angle of an equilateral triangle is } 60^\circ] \\ \implies & \angle AOP = 60^\circ && \text{ [As } \angle AOQ \text{ and } \angle AOP \text{ is the same angle.]} \space \ldots \text{(iii)} \end{aligned} ^@
  5. We know that the sum of angles of a triangle is ^@ 180^\circ. ^@

    For ^@ \triangle AOP ^@, ^@ \begin{aligned} & \angle AOP + \angle OAP + \angle APO = 180^\circ \\ \implies & 60^\circ + 90^\circ + \angle APO = 180^\circ && \text{ [Using } eq \text { (iii) and } eq \text{ (i)]} \\ \implies & \angle APO = 180^\circ - 60^\circ - 90^\circ = 30^\circ \end{aligned} ^@ Also, two tangents from an external point are equally inclined to the line segment joining the center to that point.
    So, ^@ \begin{aligned} \angle APB = 2 \angle APO = 2 \times 30^\circ = 60^\circ && \ldots \text{(iv)} \end{aligned}^@
  6. The lengths of the tangents drawn from an external point to a circle are equal.
    So, @^ \begin{aligned} & PA = PB \\ \implies & \angle PAB = \angle PBA && \text{ [Angles opposite to equal sides are equal.] } \space \ldots \text{(v)}\\ \end{aligned} @^
  7. Consider ^@ \triangle PAB ^@ @^ \begin{aligned} & \angle PAB + \angle PBA + \angle APB = 180^\circ && \text{[Sum of angles of a triangle.]} \\ \implies & \angle PAB + \angle PBA + 60^\circ = 180^\circ && \text{[Using } eq \text{ (iv)]} \\ \implies & 2 \angle PAB = 120^\circ && \text{[Using } eq \text{ (v)]} \\ \implies & \angle PAB = 60^\circ \\ \end{aligned} @^ Similarly, ^@ \angle PBA = 60^\circ. ^@
  8. As all the angles of the ^@ \triangle PAB ^@ measure ^@ 60^\circ ^@, it is an ^@ equilateral ^@ triangle.

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