Mathematics
In a ΔPQR, L and M are two points on the base QR such that ∠LPQ = ∠RQP and ∠RPM = ∠RQP. Prove that
(i) ΔPQL ∼ ΔRPM.
(ii) QL × RM = PL × PM.
(iii) PQ2 = QL × QR.
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Answer
(i) In ΔPQL and ΔRPM
∠LPQ = ∠MRP [Given]
∠LQP = ∠RPM [Given]
∴ ΔPQL ∼ ΔRPM (y A.A. axiom)
Hence, proved that ΔPQL ∼ ΔRPM.
(ii) Since, ΔPQL ∼ ΔRPM and corresponding sides of similar triangle are proportional to each other.
Hence, proved that QL × RM = PL × PM.
(iii) In ΔPQL and ΔRQP
∠LPQ = ∠QRP [Given ]
∠Q = ∠Q [Common]
∴ ΔPQL ∼ ΔRQP (By A.A. axiom)
Since, corresponding sides of similar triangle are proportional to each other.
Hence, proved that PQ2 = QR x QL.
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