Question

solve the Attachment⬆️​

Answer 1

Hand written Answer ❤️❤️

Answer 2

Given:

A PQR where RPQ = 90°

& PM ⊥IQR

To prove:

PM² = QM .MR

Proof:

In ΔPQR,

RPQ = 90°

So,Δ PQR is a right triangle

Using Pythagoras theorem in ΔPQR Hypotenuse = (Height) + (Base)

RQ² = PQ² + PR²

Now, inΔPMR,

PM1 QR

So, ∠PMR = 90

ΔPMR is a right triangle

Using Pythagoras theorem in Δ PMR Hypotenuse² = (Height)² + (Base)²

PR²= PM² + MR²

Similarly,

In Δ PMQ,

∠PMQ = 90

Δ PMR is a right triangle

Using Pythagoras theorem in A PMQ

Hypotenuse² = (Height)² + (Base)²

PQ² = PM² + MQ²

So, our equations are

RQ² = PQ² + PR² ....(1)

PR² = PM²+ MR² ....(2)

PQ²= PM² + MQ² ....(3)

Putting (2) & (3) in (1)

RQ² = PQ² + PR²

RQ² = (PM² + MQ²)+ (PM² + MR² )

RQ²= (PM² + PM²) + (MQ²+ MR²)

RQ² = 2PM² + (MQ²+ MR²)

(MQ MR)'= 2PM + (MQ² + MR² )

(As MQ + MR = RQ)

MQ + MR²+2 MQ x MR = 2PM + (MQ²+ MR²)

(MQ² + MR²)-(MQ²+ MR²)MQ x MR = 2PM

0+2 MQ x MR = 2PM²

2 MQ x MR = 2PM²

MQ x MR = PM²

PM² = MQ x MR

Hence proved