Question

of a parallelogram ABCD. DM intersects the
diagonal AC at P and AB produced at E.
Prove that : PE = 2 PD​

Answer 1

Answer:

Given: ABCD is a parallelogram. M is mid point of BC.

In △DMC and △MBE

∠DMC=∠BME (Vertically opposite angles)

∠DCM=∠MBE (Alternate angles)

MC=MB (M is mid point of BC)

Thus, △DMC≅△EMB (ASA rule)

Thus, DC=BE (By CPCT)

Now, In △DPC and △APE

∠DPC=∠APE (Vertically Opposite angles)

∠CDP=∠AEP (Alternate angles)

∠PCD=∠PAE (Alternate angles)

Thus, △DPC∼△EPA (AAA rule)

Hence,

PD

PE

=

CD

AE

PD

PE

=

CD

AB+BE

(since AB = BE = CD)

PD

PE

=2

PE=2PD