KnowledgeBoat Logo

Mathematics

In parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2AD. Prove that

(i) BE bisects ∠B

(ii) ∠AEB = a right angle.

Rectilinear Figures

30 Likes

Answer

Parallelogram ABCD is shown in the figure below:

In parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2AD. Prove that (i) BE bisects ∠B (ii) ∠AEB = a right angle. Rectilinear Figures, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

(i) AB = CD (Opposite sides of parallelogram are equal)

AD = BC (Opposite sides of parallelogram are equal)

From figure,

⇒ ∠1 = ∠2 (AE bisects ∠A)

⇒ ∠1 = ∠5 (Alternate angles are equal)

⇒ ∠2 = ∠5

⇒ AD = DE (As sides opposite to equal angles are equal)

Since, AB = 2AD and CD = AB

⇒ CD = 2AD

⇒ 2AD = DE + EC

⇒ 2AD = AD + EC

⇒ EC = AD.

Since, EC = AD and BC = AD

⇒ ∠6 = ∠4 (As angles opposite to equal sides are equal)

⇒ ∠6 = ∠3 (Alternate angles are equal)

∴ ∠3 = ∠4.

Since, ∠3 = ∠4, hence proved that BE bisects ∠B.

(ii) Let ∠1 = x and ∠3 = y.

∴ ∠2 = x and ∠4 = y

Since, AD || BC, sum of co-interior angles = 180

⇒ ∠A + ∠B = 180°

⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°

⇒ x + x + y + y = 180°

⇒ 2x + 2y = 180°

⇒ x + y = 90° ……….(1)

In △AEB,

⇒ ∠1 + ∠3 + ∠AEB = 180°

⇒ x + y + ∠AEB = 180°

⇒ 90° + ∠AEB = 180° (From i)

⇒ ∠AEB = 90°.

Hence, proved that ∠AEB = 90°.

Answered By

22 Likes


Related Questions