Question

The adjoining figure shows a parallelogram ABCD in which P is the mid point of AB and Q is tue mid point of CD. Prove that AE=EF=FC.​

Answer 1

Given: ABCD is a parallelogram. P is the mid point of CD.

Q is a point on AC such that CQ=(1/14)AC

PQ produced meet BC in R.

To prove : R is the mid point of BC

Construction : join BD in O.Let BD intersect AC in O.

Prove : O is the mid point of AC. {diagnols of parallelogram bisect each other }

Therefore OC = (1/2) AC

=> OQ = OC-CQ = (1/2)AC - (1/4)AC = (1/4)AC.

=> OQ = CQ

therefore Q is the mid point of OC.

In triangle OCD,

P is the mid point of CD and Q is the mid point of OC,

therefore PQ is parallel to OD (Mid point theorem)

=> PR is parallel to BD

In traingle BCD,

P is the midpoint of CD and PR is parallel to BD,

therefore R is the mid point of BC (Converse of mid point theorem)

hope this answer helpful u

Answer 2

Answer:

Since, PB=1/2AB

DQ=1/2DC

:.PB=DQ

Also, PB//DQ

DPBQ is a parallelogram

=DP//QB

Now in triangle ABF:

P is Mid-Point Of AB

PE//BF

:.PE bisects AF

AE=EF

Similarly,in triangle CDE:

QF bisects CE

:.EF=FC

:proved AE=EF=FC

Step-by-step explanation: