Question

pqr is a triangle in which pq=pr. S and T are point of pq and pr such that qt and rs are respectively the bisector of < pqr and < qrp, prove that ∆tqr =∆srq​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Since , Pq=Pr =] angle Q=angle R

And Qt and Rs bisects these angles

so, angle Q/2= angle R/2 =] -

thus; PS= QT -{i}

Hence; in ∆ tqr and ∆ srq; we have :

QR = RQ{ Common}

Ps=QT - {from eq. i}

angel Q = angle R { already proved}

Thus, by sas we have : ∆ tqr≈∆srq

Hence, ∆tqr= ∆ srq

Proved!

Hope it helps !