In a triangle ABC, the incircle (centre O) touches BC. CA and AB at points P, Q and R respectively, Calculate :

$∠ Q O R = 2 ∠ Q P R$
given that $∠ A = 60^{o}$

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Solution

∠AQO = ∠ARO = 90°
∠AQO + ∠ARO = 180°

Those two angles, ∠AQO and ∠ARO, are supplementary, so quadrilateral AQOR is cyclic (not on the incircle), and the other pair of opposite angles are supplementary.

∠QOR + ∠QAR = 180°
∠QOR = 180° - ∠QAR
= 180° - 60°
= 120°

Angle QPR is a circumferal angle in the incircle intercepting the same arc as central angle QOR.

∠QPR = ∠QOR/2
= 120°/2
= 60°

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In a triangle ABC, the incircle (centre O) touches BC. CA and AB at points P, Q and R respectively, Calculate : ∠QOR=2∠QPR given that ∠A=60o

In a triangle ABC, the incircle (centre O) touches BC. CA and AB at points P, Q and R respectively, Calculate :

$∠ Q O R = 2 ∠ Q P R$
given that $∠ A = 60^{o}$