In the adjoining figure, ABCD is a ||gm and E is the mid-point of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.

Question

Philoid · Accepted Answer

Here we have parallelogram ABCD with AB ∣∣ DC

Thus, DC ∣∣ BF

Now, in ∆DEC and ∆FEB,

∠DCF = ∠EBF (Alternate interior angle)

CE = BE (E is the mid-point of BC

∠CED = ∠BEF (Vertically opposite angle)

Therefore, by ASA axiom,

∆DEC ≅ ∆FEB

CD = BF (by cpct)

And CD = AB (Opposite sides of a parallelogram ABCD)

So, AF = AB + BF = AB + AB = 2AB