Question 7
In the given figure, $∠ Q > ∠ R$ , PA is the bisector of $∠ Q P R$ and PM $⊥$ QR. Prove that $∠ A P M = \frac{1}{2} (∠ Q - ∠ R)$

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Solution

PA is the bisector of $∠ Q P R$ .
$∴$ $∠ Q P A = ∠ A P R$ ....(i)
In $Δ P Q M$ $∠ Q + ∠ P M Q + ∠ Q P M = 180^{\circ}$
[Angle sum property of triangles]
$\Rightarrow ∠ Q + 90^{\circ} + ∠ Q P M = 180^{\circ}$
$[∠ P M Q = 90^{\circ}]$
$\Rightarrow ∠ Q = 90^{\circ} - ∠ Q P M$ ...(ii)
In $Δ P M R,$ $∠ P M R + ∠ R + ∠ R P M = 180^{\circ}$
[Angle sum property of triangles]
$\Rightarrow 90^{\circ} + ∠ R + ∠ R P M = 180^{\circ}$
$[∠ P M R = 90^{\circ}]$
$\Rightarrow ∠ R = 180^{\circ} - 90^{\circ} - ∠ R P M$ ... (iii)
From equations (iii) from and (ii), we get
$∠ Q - ∠ R = (90^{\circ} - ∠ Q P M) - (90^{\circ} - ∠ R P M)$
$\Rightarrow ∠ Q - ∠ R = ∠ R P M - ∠ Q P M$
$\Rightarrow ∠ Q - ∠ R = (∠ R P A + ∠ A P M) - (∠ Q P A - ∠ A P M)$ ...(iv)
$\Rightarrow ∠ Q - ∠ R = ∠ A P M + ∠ A P M$
[Using equation (1)]
$\Rightarrow ∠ Q - ∠ R = 2 ∠ A P M$
$∴ ∠ A P M = \frac{1}{2} (∠ Q - ∠ R)$

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