How i solve the problem of class 10th of triangles chapter of exercise 6.6 of ncert of question no 1. How i solve this question because i don’t know how to solve it In Figure, PS is the bisector of ∠ QPR of ∆ PQR. Prove that QS/PQ = SR/PR
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Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.
Given, PS is the angle bisector of ∠QPR. Therefore,
∠QPS = ∠SPR………………………………..(i)
As per the constructed figure,
∠SPR=∠PRT(Since, PS||TR)……………(ii)
∠QPS = ∠QRT(Since, PS||TR) …………..(iii)
From the above equations, we get,
∠PRT=∠QTR
Therefore,
PT=PR
In △QTR, by basic proportionality theorem,
QS/SR = QP/PT
Since, PT=TR
Therefore,
QS/SR = PQ/PR
Hence, proved.