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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 $AQ$.
   
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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