In fig. $6.57$ , $∠ Q > ∠ R$ . $P A$ is the bisector of $∠ Q P R$ and $P M ⊥ Q R$ .Prove that $∠ A P M = \frac{1}{2} (∠ Q - ∠ R)$ .

Open in App

Solution

Since PA is The bisector of $∠ Q P R$
$∠ Q P R = ∠ A P R$
in $△ P Q M, ∠ P Q M + ∠ P M Q + ∠ Q P M = 90^{\circ} \to (1)$
$∠ P Q M + 90 + ∠ Q P M = 180^{\circ}$ $[∵ P M ⊥ Q E \Rightarrow ∠ P M Q = 90^{\circ}]$
$∠ P Q M = 90^{\circ} - ∠ Q P M \to (2)$
In $△ P M R$
$∠ P M R + ∠ P R M + ∠ R P M = 180^{\circ}$ [by angle sum property of triangle]
$\Rightarrow 90^{\circ} + ∠ P R M + ∠ R P M = 180^{\circ}$ $[∵ P M ⊥ Q R \Rightarrow ∠ P M R = 90^{\circ}]$
$∠ P R M = 90 - ∠ R P M \to (3)$
on subtracting (3) from (2) we get
$∠ Q - ∠ R = (90 - ∠ Q P M) - (90^{\circ} - ∠ R P M)$
(where $∠ P Q M = ∠ Q$ and $∠ P R M = ∠ R)$
$∠ Q - ∠ R = ∠ R P M - ∠ Q P M$
$∠ Q - ∠ R = [∠ R P M + ∠ A P M] - [∠ Q P A - ∠ A P M] \to (4)$
$∠ Q - ∠ R = ∠ Q P A + ∠ A P M - ∠ Q P A + ∠ A P M$
$∠ Q - ∠ R = 2 ∠ A P M$
$∠ A P M = \frac{1}{2} (∠ Q - ∠ R)$ Hence proved

Suggest Corrections

Similar questions

∠ Q > ∠ R

∠ Q P R

P M ⊥ Q R

∠ A P M = 1 / 2 (∠ Q - ∠ R)

∠ Q > ∠ R

∠ Q P R

⊥

∠ A P M = \frac{1}{2} (∠ Q - ∠ R)

∠ Q > ∠ R

∠ Q P R

⊥

∠ A P M = \frac{1}{2} (∠ Q - ∠ R)

In triangle PQR, Q > angle R; PA is the bisector of angle QPR ; PM IS perpendicular to PR .prove that angle APM=1/2(angle Q - angle R).

In ∆PQR, PR > PQ, PA is the bisector of angle P and PM is perpendicular to QR. Prove that angle APM = 1/2 ( angle Q - angle R).

Join BYJU'S Learning Program