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Question -

In the fig.6.36,QR/QS = QT/PR and 1 = 2. Show that ΔPQS ~ ΔTQR.



Answer -

Solution:

In ΔPQR,

PQR = PRQ

PQ = PR ………………………(i)

Given,

QR/QS = QT/PRUsing equation (i), we get

QR/QS = QT/QP……………….(ii)

In ΔPQS and ΔTQR, by equation (ii),

QR/QS = QT/QP

Q = Q

ΔPQS ~ ΔTQR [By SAS similarity criterion]

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