In parallelogram ABCD, the angle bisector of ∠A bisects BC. Will angle bisecto

Question

In parallelogram ABCD, the angle bisector of

∠ A

bisects BC. Will angle bisector of B also bisect AD? Give reason.

Answer:
:
Given, ABCD is a parallelogram, bisector of

∠ A

, bisects BC at F, i.e.

∠ 1 = ∠ 2

, CF = FB.
Draw

F E ∥ B A

ABFE is a parallelogram by construction [

∵ F E ∥ B A

]

\Rightarrow ∠ 1 = ∠ 6

[alternate angle]

In Parallelogram ABCD, The Angle Bisector Of ∠A Bisects BC...

But

∠ 1 = ∠ 2

[given]

∴ ∠ 2 = ∠ 6

AB = FB [opposite sides to equal angles are equal] ..... (i)

∴ A B F E i s a r h o m b u s

.
Now, in

Δ A B O

and

Δ B O F

, AB = BF [from Eq. (i)]
BO = BO [common]
AO = FO [diagonals of rhombus bisect each other]

∴ Δ A B O ≅ Δ B O F

[by SSS]

∠ 3 = ∠ 4

[by CPCT]
Now,

B F = \frac{1}{2} B C

[given]

\Rightarrow B F = \frac{1}{2} A D

[BC = AD]

\Rightarrow A E = \frac{1}{2} A D

[BF = AE]

∴

E is the mid-point of AD.

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