In Fig 7.51, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.

 

Solution:

It is given that PR > PQ and PS bisects ∠QPR

Now we will have to prove that angle PSR is smaller than PSQ i.e. ∠PSR > ∠PSQ

Proof:

∠QPS = ∠RPS — (ii) (As PS bisects ∠QPR)

∠PQR > ∠PRQ — (i) (Since PR > PQ as angle opposite to the larger side is always larger)

∠PSR = ∠PQR + ∠QPS — (iii) (Since the exterior angle of a triangle equals to the sum of opposite interior angles)

∠PSQ = ∠PRQ + ∠RPS — (iv) (As the exterior angle of a triangle equals to the sum of opposite interior angles)

By adding (i) and (ii)

∠PQR +∠QPS > ∠PRQ +∠RPS

Thus, from (i), (ii), (iii) and (iv), we get

∠PSR > ∠PSQ