Question

3. In trapezium ABCD, AB is parallel to DC; P and Q are the mid-points of AD and BC
respectively. BP produced meets CD produced to point E. Prove that:

a. point P bisects BE.
b. PQ is parallel to AB.​

Answer 1

Step-by-step explanation:

Given: ABCD is a trapezium. AB∥DC. P and Q are mid points of AD and BC respectively.

BP produced meets CD at E

To prove: p is mid point of BE.

In △APB and △EPD

∠APB=∠EPD ...(Vertically opposite angles)

∠EDP=∠PAB ...(Alternate angles)

PA=PD ...(P is mid point of AD)

Thus, △APB≅△DPE ...(ASA rule)

Hence, PE=PB ...(By cpct)

thus, P is mid point of BE.

for traingle ecb, pq|| ce

ce || ab

then, pq||ab

proved