Question

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.

Answer 1

ABCD is a parallelogram.
i.e., AB ∣∣ DC
∴​ DC ∣∣​ BF
In ∆DEC and ∆FEB, we have:
EC = EB (E is the mid point of BC)
∠CED = ∠BEF (Vertically opposite angles)
∠ECD = ∠EBF (Alternate interior angles)
∴​ ∆​DEC ≅ ∆FEB
​⇒ CD = BF (CPCT)
Also, CD = AB (Opposite sides of a parallelogram)
Now, AF = AB + BF
⇒ AF = AB + AB
∴ AF = 2AB ​