Question

ABCD is a parallelogram. AB is produced to E so that BE = AB. A line through E drawn parallel to AC meets CB produced at F. Show that ACEF is a parallelogram.

[Hint: Show that ΔABC ≅ ΔEBF ∴ BC = BF]

Answer 1

Given:

• ABCD is a parallelogram.
• AB is produced to E such that BE = AB.
• A line through E is drawn parallel to AC and meets CB produced at F.

To prove: ACEF is a parallelogram.

Proof (stepwise):

1. Since ABCD is a parallelogram, opposite sides are equal: AB = DC, AD = BC

2. Given BE = AB, so BE = AB = DC ...(From step 1)

3. Consider triangles ΔABC and ΔEBF:

   • AB = BE   ...(Given)
   • ∠ABC = ∠EBF   ...(Since line through E is parallel to AC, alternate interior angles are equal)
   • BC = BF will be shown by congruence

4. Using SAS Side-Angle-Side criterion:

   In ΔABC and ΔEBF, AB = BE, ABC = ∠EBF, BC = BF. So, ΔABC ≅ ΔEBF which implies BC = BF.

5. Since EF || AC given as E through a line parallel to AC.

6. Also, AE is a straight line extending AB. So, AE || CF because CF lies along CB produced and since BC = BF, CF is effectively a translated line parallel to AE.

7. Thus, in quadrilateral ACEF: AE || CF and EF || AC.

8. A quadrilateral with both pairs of opposite sides parallel is a parallelogram.

Since AE || CF and EF || AC, quadrilateral ACEF is a parallelogram.