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# In figure, PR > PQ and PS bisects $$\angle{QPR}$$. Prove that $$\angle{PSR}$$ > $$\angle{PSQ}$$. In $$\triangle{PQR}$$, we have,
PR > PQ ...(given)
$$\therefore$$ $$\angle{PQR}$$ > $$\angle{PRQ}$$ ...(i)
($$\because$$ angle opposite to longer side is greater)
We can also say,
$$\angle{a}$$ = $$\angle{b}$$ ...(ii)
($$\because$$ PS bisects $$\angle{QPR}$$) On adding Equation (i) and (ii), we get,

$$\angle{PQR}$$ + $$\angle{a}$$> $$\angle{PRQ}$$ + $$\angle{b}$$ ...(iii)
Also, $$\angle{PQS}$$ + $$\angle{PSQ}$$ + $$\angle{a}$$ = $$180^\circ$$ ...(iv)
and $$\angle{PRS}$$ + $$\angle{PSR}$$ + $$\angle{b}$$ = $$180^\circ$$ ...(v)
($$\because$$ the sums of all angles of a triangle are $$180^\circ$$)

From Equation (iv) and (v), we get,

$$\angle{PQS}$$ + $$\angle{PSQ}$$ + $$\angle{a}$$ = $$\angle{PRS}$$ + $$\angle{PSR}$$ + $$\angle{b}$$
$$\because$$ $$\angle{PQS}$$ = $$\angle{PQR}$$ and $$\angle{PRS}$$ = $$\angle{PRQ}$$,
we get,

$$\angle{PQR}$$ + $$\angle{PSQ}$$ + $$\angle{a}$$ = $$\angle{PRQ}$$ + $$\angle{PSR}$$ + $$\angle{b}$$

Rearranging the terms, we get,

$$\angle{PQR}$$ + $$\angle{a}$$ + $$\angle{PSQ}$$ = $$\angle{PRQ}$$ + $$\angle{b}$$ + $$\angle{PSR}$$ ...(vi)

From Equation (iii) and (vi), we get,

$$\angle{PSQ}$$ < $$\angle{PSR}$$ ...
($$\because$$ side opposite to greater angle is longer)
Hence, proved.