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  • In Fig. 6.17, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR. Prove that ∠APM = 1/2 (∠Q – ∠R).

In Fig. 6.17, \angle Q > \angle R, PA is the bisector of \angle QPR and PM \perp QR. Prove that  \angle APM = \frac{1}{2} (\angle Q -\angle R).

Answers (1)

Solution:

Given : In \triangle PQR, \angle Q > \angle R

PA is the bisector of \angle QPR  and PM \perp QR.

To prove : \angle APM =\left ( \frac{1}{2} \right ) (\angle Q -\angle R)
Proof: Since PA is the bisector of \angle P, we have

\angle APQ = \left ( \frac{1}{2} \right )\angle P                        … (i)
In right angled \triangle PMQ we have

\angle Q + \angle MPQ + 90^{\circ} = 180^{\circ} (Angle sum property)

\Rightarrow \angle MPQ = 90^{\circ} - \angle Q              … (ii)
Now 
\angle APM = \angle APQ - \angle MPQ
\angle APM =\frac{1}{2} \angle P - (90 - \angle Q)         using (i) & (ii)
\angle APM = \frac{1}{2} \angle P - 90^{\circ} + \angle Q
\angle APM = \frac{1}{2}\angle P -\frac{1}{2} (\angle P + \angle R + \angle Q) + \angle Q  Since 90^{\circ} = \frac{1}{2}(\angle P + \angle R + \angle Q)
\angle APM = \frac{1}{2}\angle P -\frac{1}{2} \angle P - \frac{1}{2} \angle R-\frac{1}{2} \angle Q + \angle Q
\angle APM =\frac{1}{2} (\angle Q -\angle R)

Hence proved

 

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