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

In the given figure, ABCD is a quadrilateral. Diagonal BD bisects ∠B and ∠D both.

Ans:

(i) Let’s start with Δ ABD:

∠ 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. In the given figure, ABCD is a quadrilateral. Diagonal BD bisects ∠B and ∠D both.

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)

\frac{AB}{BC} = \frac{AE}{CE} (by angle bisector theorem)

\frac{AB}{BC} = 1 ( ∵ AE = CE)

∴ AB = BC

Hence Proved !

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