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Question

Prove that the tangent at the extremities of any chord makes equal angles at the chord. 


Solution


Let $$AB$$ be a chord of a circle with centre $$O$$, and let $$AP$$ and $$BP$$ be the tangents at $$A$$ and $$B$$ respectively. 

Suppose the tangents meet at $$P$$. Join $$OP$$. Suppose $$OP$$ and meets $$AB$$ at $$C$$. 

We have to prove that $$\angle PAC=\angle PBC$$.

In triangle $$PCA$$ and $$PCB$$, we have 

$$\Rightarrow$$$$PA=PB\quad$$          [$$\because\quad$$ tangents from an external point are equal]
$$\Rightarrow$$$$\angle APC=\angle BPC\quad$$  [$$\because\quad PA$$ and $$PB$$ are equal inclined to $$OP$$].
$$\Rightarrow$$$$PC=PC$$

So, by $$SAS------$$criterion of congruence, we have

     $$\triangle PAC\cong \triangle PBC$$
$$\Rightarrow \angle PAC= \angle PBC$$ 

1031485_1009631_ans_5558c2aaf92345919e94286877aa590a.png

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