# In the figure, PS is the bisector of $$\triangle$$ PQR. Prove that $$QS\over SR$$ = $$PQ\over PR$$.

## Solution :

Given : PQR is a triangle and PS is the bisector of $$\angle$$ QPR meeting QR at S. $$\therefore$$  $$\angle$$ QPS = $$\angle$$ SPR

To Prove : $$QS\over SR$$ = $$PQ\over PR$$

Construction : Draw RT parallel to SP to cut QP produced at T.

Proof : Since PS || TR and PR cuts them, hence we have :

$$\angle$$ SPR = $$\angle$$ PRT      …….(1)       (alternate angles) and $$\angle$$ QPS = $$\angle$$ PTR         ………(2)     (corresponding angles)

but,  $$\angle$$ QPS = $$\angle$$ PTR        ………(3)      (given)

$$\therefore$$  $$\angle$$ PRT = $$\angle$$ PTR       [From (1) and (2)]

$$\implies$$  PT = PR           ……….(3)        (sides opposite to equal angles are equal)

Now, in $$\triangle$$ QRT, we have :

SP || RT               (by construction)

$$\therefore$$  $$QS\over SR$$ = $$PQ\over PT$$       (by basic proportionality)

or          $$QS\over SR$$ = $$PQ\over PR$$           (from 3)