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In∆PQR,<1=<2 [Given]
PR=PQ [side opposite to equal<of a triangle are also equal]
Given that QR/QS =QT/PR
QR/QS=QT/PQ
therefore PQ=PR Proved above
QS/QR =PQ/QT-(i) [taking] Reciprocal]
In∆PQS and ∆TQR,
<PQS =<TQR [ Common]
QS/QR =QP/QT[from eq(i)]
therefore ∆PQS~∆TQR[By SAS]
Hence it is proved
PR=PQ [side opposite to equal<of a triangle are also equal]
Given that QR/QS =QT/PR
QR/QS=QT/PQ
therefore PQ=PR Proved above
QS/QR =PQ/QT-(i) [taking] Reciprocal]
In∆PQS and ∆TQR,
<PQS =<TQR [ Common]
QS/QR =QP/QT[from eq(i)]
therefore ∆PQS~∆TQR[By SAS]
Hence it is proved
riya42150:
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OK I'll do it later
ok byee bro
ur wlcm
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