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# In figure,$$\angle{PQR}$$ = $$\angle{PRQ}$$, then prove that$$\angle{PQS}$$ = $$\angle{PRT}$$ .

By Linear pair axiom,
$$\angle{PQS}$$ + $$\angle{PQR}$$ = $$180^\circ$$...(i)

Similarly, again by Linear pair axiom,
$$\angle{PRQ}$$ + $$\angle{PRT}$$ = $$180^\circ$$ ...(ii)

Thus by Eq. (i) and (ii),we get,
$$\angle{PQS}$$ + $$\angle{PQR}$$ = $$\angle{PRQ}$$ + $$\angle{PRT}$$
As it is given, $$\angle{PQR}$$ = $$\angle{PRQ}$$
$$\therefore$$ $$\angle{PQS}$$ + $$\angle{PRQ}$$ = $$\angle{PRQ}$$ + $$\angle{PRT}$$
By cancelling equal terms,
we get, $$\angle{PQS}$$ = $$\angle{PRT}$$
Hence, proved.