Question

In the figure ABCD is a parallelogram and E is the midpoint of side BC. DE and AB on producing meet at F. Prove that AF =2AB.

Answer 1

ABCD is a parallelogram. E is the midpoint of BC. So, BE =CE.
DE produced meets the AB produced at F.
Consider the triangles CDE and BFE.
BE =CE [Given]
$\angle CED=\angle BEF$ [ Vertically opposite angles ]
$\angle DCE=\angle BEF$ [Alternate angles]
$\therefore \Delta CDE\cong \Delta BFE$
So, CD = BF [CPCT]
But, CD = AB
Therefore, AB = BF
AF = AB + BF
AF = AB + AB
AF = 2AB