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Q2 PQR is a triangle right angled at P and M is a point on QR such that $PM \perp QR$ . Show that 5 $PM ^2 = QM . MR .$

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Let $\angle MPR$ be x

In $\triangle MPR$ ,

$\angle MRP=180 \degree-90 \degree-x$

$\angle MRP=90 \degree-x$

Similarly,

In $\triangle MPQ$ ,

$\angle MPQ=90 \degree-\angle MPR$

$\angle MPQ=90 \degree-x$

$\angle MQP=180 \degree-90 \degree-(90 \degree-x)=x$

In $\triangle QMP\, and\, \triangle PMR,$

$\angle MPQ\, =\angle MRP$

$\angle PMQ\, =\angle RMP$

$\angle MQP\, =\angle MPR$

$\triangle QMP\, \sim \triangle PMR,$ (By AAA)

$\frac{QM}{PM}=\frac{MP}{MR}$

$\Rightarrow PM^2=MQ\times MR$

Hence proved

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