in quadrilateral PQRS the bistector of angle R and angle S meets at point T . show that angle P + angle Q =2 angle RTS
in quadrilateral PQRS the bistector of angle R and angle S meets at point T . show that angle P + angle Q =2 angle RTS
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We have a quadrilateral PQRS , where bisectors of angle R and angle S meet at point T.
We know in quadrilateral,
∠ P + ∠Q + ∠R + ∠S = 360°
∠ P + ∠Q = 360° - (∠R + ∠S) --------------- (1)
In ∆RTS ,
∠RTS + ∠TSR + ∠SRT = 180° --------------- (2)
Given,
∠TSR = ∠S/2
∠SRT = ∠R/2
After substitute this in equation (2) , we get
∠RTS + ∠S/2 +∠R/2 = 180°
⇒ 2∠RTS +∠S +∠R = 360°
⇒ 2∠RTS = 360° - (∠R + ∠S)
From equation number (1)
⇒ 2∠RTS = ∠ P + ∠Q (Hence proved)
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