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 

Answer 1

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