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Question

In the adjoining figure, $$OPQR$$ is a square. A circle drawn with centre $$O$$ cuts the square in $$X$$ and $$Y$$.
Prove that $$QX=QY$$

Solution
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Consider $$\triangle OXP$$ and $$\triangle OYR$$

We know that $$\angle OPX$$ and $$\angle ORY$$ are right angles

So we get

$$\angle OPX=\angle ORY =90^{o}$$

We know that $$OX$$ and $$OY$$ are the radii

$$OX=OY$$

From the figure, we know that the sides of a square are equal

$$OP=OR$$

By RHS congruence criterion

$$\triangle OXP\cong OYR$$

$$PX=RY (c.p.c.t.)$$

We know that $$PQ=QR$$

So we get

$$PQ-PX=QR-RY$$

$$QX=QY$$

Therefore, it is proved that $$QX=QY$$

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