Question

In fig 14.36, ABCD is a parallelogram and E is the mid-point of side BC. IF DE and AB when produced meet at F, prove that AF = 2AB.

Answer 1

Solution :-

in the figure

$△DCE$ and BFE

any DEC = any BEF (vertically opp any)

$EC=BE$ ($E$ is the mid point)

$\angle DCB=\angle EBF$ (alternate angle DC parallel to AF)

So $△DCE$ congruent to $△BFE$

Therefore $DC=BF$ ...(1)

now CD = AB (ABCD is a parallelogram)

$soAF=AB+BF$

$=AB+DC$ from $\left(1\right)$

$=AB+AB$

$=2AB$