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Question
In ΔPQR, L and M are two points on the base QR, such that ∠LPQ = ∠QRP and ∠RPM = ∠RQP.
Prove that : (i) ΔPQL ∼ ΔRPM
(ii) QL. Rm = PL. PM
(iii) PQ2 = QR. QL.
Solution
(i) Consider
ΔPQL and ΔRPM
Since ∠PQL = ∠RPM
and ∠QPL = ∠RPM and
∠QPL = ∠PRM
By A. A. Criterion
ΔPQL ∼ ΔRPM
Hence Proved.
(ii) In ΔPQL ∼ ΔRPM
`"PQ"/"RP" = "OL"/"PM" = "PL"/"MR"`
then `"OL"/"PM" = "PL"/"MR"`
QL · MR = PL · PM.
Hence Proved.
(iii) In ΔPQR and ΔLQP
∠PQR = ∠LQP
PQ = PQ
Hence ΔPQR ∼ ΔLQP
`"QR"/"PQ" = "PQ"/"LQ"`
PQ2 = QR · QL.
Hence proved.
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