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In the figure, PS is the bisector of  of . Prove that

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Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T. Given that, PS is the angle bisector of ∠QPR. ∠QPS = ∠SPR … (1) By construction, ∠SPR = ∠PRT (As PS || TR) … (2) ∠QPS = ∠QTR (As PS || TR) … (3) Using these...
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Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given that, PS is the angle bisector of ∠QPR.

∠QPS = ∠SPR … (1)

By construction,

∠SPR = ∠PRT (As PS || TR) … (2)

∠QPS = ∠QTR (As PS || TR) … (3)

Using these equations, we obtain

∠PRT = ∠QTR

∴ PT = PR

By construction,

PS || TR

By using basic proportionality theorem for ΔQTR,

  

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