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Solve this:

20) In the figure, PQ$\text{\u22a5}$QR and PM $\text{\u22a5}$RM. Also PQ=PM, show that QR=MR.

Please find below the solution to the asked query:

In $\u25b3$ PQR and $\u25b3$ PMR

PQ = PM ( Given )

$\angle \mathrm{PQR}=\angle \mathrm{PMR}\hspace{0.17em}=90\xb0(\mathrm{Given}\mathrm{PQ}\perp \mathrm{QR}\mathrm{and}\mathrm{PM}\perp \mathrm{MR}\hspace{0.17em})$

And

PR = PR ( Common side ) ( That is also hypotenuse for both triangles )

So,

$\u25b3$ PQR $\cong $ $\u25b3$ PMR ( By RHL rule )

Then,

**QR = MR ( By CPCT ) ( Hence proved )**

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