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Question

In the given figure, $$XY$$ and $$X'Y'$$ are two parallel tangents to a circle with centre $$O$$ and another tangent $$AB$$ with point of contact $$C$$ intersecting $$XY$$ at $$A$$ and $$X'Y'$$ at $$B$$. Prove that $$\angle AOB = 90^{\circ}$$.
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Solution

In $$\triangle AOP$$ & $$\triangle AOC$$
$$OP=OC$$                          [Both radius]
$$AP=AC$$                          [Length of tangents drawn from external point to a circle are equal]
$$OA=OA$$                          [common]
$$\therefore \triangle AOC \cong \triangle AOP$$       [SSS Congruence rule]
So, $$\angle AOP=\angle AOC...............(1)$$                          [CPCT]
Now,
In $$\triangle BOC$$ & $$\triangle BOQ$$
$$OC=OQ$$                         [Both radius]   
$$BC=BQ$$                          [Length of tangents drawn from external point to a circle are equal]
$$OB=OB$$                          [common]
$$\therefore \triangle BOC \cong \triangle BOQ$$       [SSS Congruence rule]
So, $$\angle BOC=\angle BOQ...............(2)$$                          [CPCT]
For line PQ
$$\angle AOP+\angle AOC+\angle BOC+\angle BOQ=180$$
$$\angle AOC+\angle AOC+\angle BOC+\angle BOC=180$$
$$2\angle AOC+2\angle BOC=180$$
$$2(\angle AOC+\angle BOC)=180$$
$$\angle AOC+\angle BOC=\frac{180}{2}$$
$$\angle AOC+\angle BOC=90$$
$$\angle AOB=90$$
Hence proved.

Maths

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