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In ΔPQR, ∠PQR = ∠PRQ
∴ PQ = PR ...(i)
Given,QR/QS = QT/PR
Using (i), we get
QR/QS = QT/QP ...(ii)
In ΔPQS and ΔTQR,
QR/QS = QT/QP [using (ii)]
∠Q = ∠Q
∴ ΔPQS ~ ΔTQR [SAS similarity criterion]
∴ PQ = PR ...(i)
Given,QR/QS = QT/PR
Using (i), we get
QR/QS = QT/QP ...(ii)
In ΔPQS and ΔTQR,
QR/QS = QT/QP [using (ii)]
∠Q = ∠Q
∴ ΔPQS ~ ΔTQR [SAS similarity criterion]
mrinmai1:
welcome ra
please help me
u r frm class 10 isnt it
ha
see my questions in my profile and answer the first one
yo muzine
tangent one u need help
isnt it
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