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 $$\angle AOB = 90^{\circ}$$.

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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