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
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Answer
Parallelogram ABCD is shown in the figure below:
(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°.
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