Question

PQR is a triangle right angled at P and M is a point on QR such that PM perpedicular QR .Show that PM²= OM.MR​

Answer 1

Answer:

In △PMR,

By Pythagoras theorem,

(PR)

2

=(PM)

2

+(RM)

2

.......(1)

In △PMQ,

By Pythagoras theorem,

(PQ)

2

=(PM)

2

+(MQ)

2

.......(2)

In △PQR,

By Pythagoras theorem,

(RQ)

2

=(RP)

2

+(PQ)

2

........(3)

∴ (RM+MQ)

2

=(RP)

2

+(PQ)

2

∴ (RM)

2

+(MQ)

2

+2RM.MQ=(RP)

2

+(PQ)

2

....(4)

Adding 1) and 2) we get,

(PR)

2

+(PQ)

2

=2(PM)

2

+(RM)

2

+(MQ)

2

...(5)

From 4) and 5) we get,

2RM.MQ=2(PM)

2

∴ (PM)

2

=RM.MQ