Question

A trapezium abcd in which CD is parallel to ab prove that ap/dc=bp/dq

Answer 1

Answer:Given, a trapezium ABCD in which AB || DC. P and Q are points on AD and BC, respectively such that PQ || DC. Thus, AB || PQ || DC.

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: E 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.

Hope it helped u