In the following figure, M is the mid-point of BC of a parallelogram ABCD. DM intersects the diagonal AC at P and AB produced at E. Prove that : PE = 2PD.

In ∆BME and ∆DMC,

∠BME = ∠CMD [Vertically opposite angles are equal.]

∠MCD = ∠MBE [Alternate angles are equal]

BM = MC [M is mid-point of BC]

∴ ∆BME ≅ ∆DMC [By AAS congruence rule]

∴ BE = DC [By C.P.C.T]

Since, opposite sides of parallelogram are equal.

∴ AB = DC

or, AB = DC = BE. ………..(1)

In ∆DCP and ∆EPA,

∠DPC = ∠EPA [Vertically opposite angles are equal.]

∠CDP = ∠AEP [Alternate angles are equal]

∴ ∆DCP ~ ∆EAP [By AA]

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{DC}{EA} = \dfrac{PD}{PE} \\[1em] \Rightarrow \dfrac{EA}{DC} = \dfrac{PE}{PD} \\[1em] \Rightarrow \dfrac{PE}{PD} = \dfrac{AB + BE}{DC} \\[1em] \Rightarrow \dfrac{PE}{PD} = \dfrac{2DC}{DC} \\[1em] \Rightarrow \dfrac{PE}{PD} = 2 \\[1em] \Rightarrow PE = 2PD.$

Hence proved that PE = 2PD.