Q) In the given figure, ABCD is a quadrilateral. Diagonal BD bisects ∠ B and ∠ D both. Prove that:
(i) Δ ABD ~  Δ CBD
(ii) AB = BC

Ans:

∠ BAD + ∠ ABD + ∠ BDA = 180 …. (i)

Similarly in Δ BCD:

∠ BCD  + ∠ DBC + ∠ BDC= 180 ……… (ii)

Comparing equations (i) and (ii), we get:

∠ BAD + ∠ ABD + ∠ BDA = ∠ BCD + ∠ DBC + ∠ BDC ……… (iii)

We are given that BD is bisector of ∠ B and ∠ D both

∴ ∠ ABD = ∠ DBC and ∠ BDA = ∠ BDC

substituting these values in equation (iii), we get:

∠ BAD + ∠ ABD + ∠ BDA = ∠ BCD + ∠ ABD + ∠ BDA

∴ ∠ BAD = ∠ BCD

Next, let’s compare Δ  ABD and Δ  CBD:

∠ ABD = ∠ DBC  (BD is angle bisector of ∠ B)

∠ BDA = ∠ BDC (BD is angle bisector of ∠ D)

∠ BAD = ∠ BCD (calculated above)

∴ Δ  ABD ~ Δ  CBD

Hence Proved!

(ii) Construction: let’s connect diagonal AC.

It cuts BD at point E.

Let’s look into Δ  ABE and Δ  CBE

∠ ABE = ∠ EBC  (BD and hence BE is angle bisector of ∠ B)

∴ AE = CE (sides opposite to equal angles are equal)

(by angle bisector theorem)

= 1 ( ∵ AE = CE)

∴ AB = BC

Hence Proved !

Please do press “Heart” button if you liked the solution.

Scroll to Top