Question

In △PQR, MN⊥PQ, MT⊥PR, & MN=MT. If∠QPR=70°, What type of cevian is PM?Also, find∠PMT.

Answer 1

Answer

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

Step-by-step explanation:

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