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Question
P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that: DP : PL = DC : BL.
Solution
In ∆DPC and ∆BPL, we have
∠DPC = ∠BPL ...[Vertically opposite angles are equal]
∠DCP = ∠PBL ...[Alternate angles (as AB || DC) are equal]
∴ ∆DPC ~ ∆BPL ...[By A.A.]
Since the corresponding sides of similar triangles are proportional.
`(DP)/(PL) = (DC)/(BL)`
i.e., DP : PL = DC : BL
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