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5.  In Fig ,  \small PR>PQ  and PS bisects \small \angle QPR. Prove that  \small \angle PSR>\angle PSQ.


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We are given that     \small PR>PQ.

Thus                 \angle PQR\ =\ \angle PRQ

Also,  PS bisects \small \angle QPR , thus :

                          \angle QPS\ =\ \angle RPS

Now, consider \Delta QPS

                         \angle PSR\ =\ \angle PQR\ +\ \angle QPS                (Exterior angle)

Now, consider \Delta PSR

                         \angle PSQ\ =\ \angle PRQ\ +\ \angle RPS

Thus from the above the result we can conclude that :

                                 \small \angle PSR>\angle PSQ

Posted by

Sanket Gandhi

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