Question

ABCD is a parallelogram. M is the midpoint of BC. Show that DC = CP. [Hint: Prove that triangleABM = trianglePCM]

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Answer 1

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,$\frac{PE }{PD} = \frac{AE}{CD}$

$\frac{PE }{PD}=\frac{AB+BE}{CD}$

(since AB = BE = CD)

$\frac{PE }{PD} = 2$

PE=2PD