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Question
In the following figure, M is mid-point of BC of a parallelogram ABCD. DM intersects the diagonal AC at P and AB produced at E. Prove that : PE = 2 PD
Solution
In ΔBME and ΔDMC,
∠BME = ∠CMD ...(Vertically opposite angles)
∠MCD = ∠MBE ...(Alternate angles)
BM = MC ...(M is the mid-point of BC)
So, ΔBME ≅ ΔDMC ...(AAS congruence criterion)
`=>` BE = DC = AB
In ΔDCP and ΔEPA,
∠DPC = ∠EPA ...(Vertically opposite angles)
∠CDP = ∠AEP ...(Alternate angles)
ΔDCP ∼ ΔEAP ...(AA criterion for similarity)
`=> (DC)/(EA) = (PC)/(PA) = (PD)/(PE)`
`=> (EA)/(DC) = (PE)/(PD)`
`=> (PE)/(PD) = (AB + BE)/(DC)`
`=> (PE)/(PD) = (2DC)/(DC)`
`=> (PE)/(PD) = 2`
`=>` PE = 2PD
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