In a right angle triangle pqr …

Given: PQR is a triangle right angled at P and M is a point on QR such that PM {tex} \bot {/tex} QR
 
To prove PM² = QM.MR
Proof: In right triangle PQR,
QR² = PQ² + PR² (1).............[By Pythagoras theorem]
In right triangle PMQ,
PQ² = PM² + MQ² (2)...........[By Pythagoras theorem]
In right triangle PMR,
PR² = PM² + MR² (3) ............[By Pythagoras theorem]
Using (2) and (3), (1) gives
QR² = (PM² + MQ²) + (PM² + MR²)
{tex}\Rightarrow {/tex} QR² = 2PM² + MQ² + MR²
=(MQ + MR)² = 2PM² + MQ² + MR²
{tex}\Rightarrow {/tex} MQ² + MR² + 2MQ.MR
= 2PM² + MQ² + MR²
{tex}\Rightarrow {/tex} PM² = QM.MR
A litre.
In {tex}\triangle {/tex}QMP and {tex}\triangle {/tex}PMR,
{tex}\angle{/tex} QMP= {tex}\angle{/tex} PMR......[Each equal to 90^o]
{tex}\angle{/tex}MQP {tex} \sim {/tex} {tex}\angle{/tex}PMR.......AA similarity criterion
{tex}\therefore \frac{{QM}}{{PM}} = \frac{{PM}}{{RM}}{/tex} {tex}\therefore {/tex} corresponding sides of two similar triangles
{tex}\Rightarrow {/tex} PM² = QM.RM
{tex}\Rightarrow {/tex} QM.MR