Question

In a triangle PQR, PR2-PQ2 ​=QR2 and M is a point on side PR such that QM is perpendicular to PR. Prove that QM2=PM*MR

Answer 1

[FIGURE IS IN THE ATTACHMENT]
Given:
In ∆ PQR, PR²-PQ²= QR² & QM ⊥ PR
To Prove: QM² = PM × MR
Proof:
Since, PR² - PQ²= QR²
PR² = PQ² + QR²
So, ∆ PQR is a right angled triangle at Q.
In ∆ QMR & ∆PMQ
∠QMR = ∠PMQ [ Each 90°]
∠MQR = ∠QPM [each equal to (90°- ∠R)]
∆ QMR ~ ∆PMQ [ by AA similarity criterion]
By property of area of similar triangles,
ar(∆ QMR ) / ar(∆PMQ)= QM²/PM²
1/2× MR × QM / ½ × PM ×QM = QM²/PM²
[ Area of triangle= ½ base × height]
MR / PM = QM²/PM²
QM² × PM = PM² × MR
QM² =( PM² × MR)/ PM
QM² = PM × MR

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Answer 2

hope this helps you