Please help me with this sum...based on Pythagoras theorem?

Question

Given a triangle PQR, right angled at P. PM is a perpendicular drawn from P to hypotenuse QR. Prove that PR^2 = PQ^2 + QR^2 - 2 QM x QR

Answer 1

Triangle PQM and Triangle PQR are similar
[(Angles are equal. Sides are unequal but proportional) >> one right angle in each triangle and angle Q is common]'

PQ / QR = QM / PQ
Cross multiplying
QM x QR = PQ^2 .... ... (1)

Right side
= PQ^2 + QR^2 - 2 QM x QR
= PQ^2 + QR^2 - 2 PQ^2 [from 1]
= QR^2 - PQ^2
= PR^2 [PQR is right-angled triangle]
= Left side

Answer 2

I got the following identities:
QR²=PQ²+PR²
PQ²=QM²+MP²
PR²=MP²+MR²

and

QR² = (QM + MR)²

Work those to get the equation above.

Answer 3

look at the stuff in what you have to prove.
ie. you need PR2, PQ2, QR2, etc..
and then identify the triangles that they come in, and apply pythagoras theorem to each of those triangles.
then see what you NEED again, and add or subtract the various equations. or equate values, if something is repeated..
its an easy sum... good luck !