Question

$PQR$ is a triangle right angled at $P$ and $M$ is a point on $QR$ such that $PM\perp QR$.

Show that $P{M}^2=QM.MR$

• A. $P{M}^2=QM.MR$
• B.
• C.
• D.

Answer 1

Answer: (A)

Given:

In $△PQR,\angle P={90}^{\circ }$

$PM\perp QR$

To prove: $P{M}^2=QM.MR$

Proof:

In $△PQR$,

Let $\angle Q=\theta$

$\Rightarrow \angle R={90}^{\circ }-\theta$ [Since, angle sum property of triangle]

Now, it is given that, $PM\perp QR$

$\Rightarrow$ In $△PMR$,

$\angle RMP={90}^{\circ }$

$\angle RPM={90}^{\circ }-\theta$ [Since, angle sum property of triangle]

Similarly in $△PMQ$

$\angle PQM=\theta$ [Assumed]

$\angle QMP={90}^{\circ }$

$\angle QPM={90}^{\circ }-\theta$

Now, In $△PQM$ and $△RPM$

$\angle PQM=\angle RPM=\theta$

$\angle PMQ=\angle PMR={90}^{\circ }$

$\Rightarrow$ By $A.A$ Similarity criterion,

$△PQM\sim △RPM$

$\Rightarrow$ Corresponding sides are in the same ratio.

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

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

$\Rightarrow P{M}^2=QM.MR$

Hence, proved.