A triangle PQR where PQ=QR and angle Q=60 is inscribed in a circle . If S is any point in minor arc QR but not coincide with Q and R then prove that PS bisects angle QSR
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PS bisects ∠QSR in A triangle PQR where PQ=QR and angle Q=60 is inscribed in a circle
Step-by-step explanation:
A triangle PQR where PQ=QR and angle Q=60
PQ = QR
=> ∠Q = ∠R
=> ∠R = 60°
∠P + ∠Q + ∠R = 180°
=> ∠P = 60°
Hence PQR is an equilateral triangle
=> PQ = PR = QR
PS will create two angle
∠PSQ & ∠PSR
if PS bisects ∠QSR
then ∠PSQ = ∠PSR
∠PSQ is angle subtended by chord PQ at arc segment
∠PSR is angle subtended by chord by PR at arc segment
PQ = PR
Angle subtended by equal length chord are equal
hence ∠PSQ = ∠PSR
Hence PS bisects ∠QSR
Learn more:
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