Mathematics
In the given figure, ABCD is a parallelogram. Prove that : AB = 2BC.
Rectilinear Figures
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Answer
Given,
ABCD is a parallelogram.
∴ AB || CD (Opposite sides of parallelogram are parallel)
From figure,
AE is the transversal.
⇒ ∠BAE = ∠AED (Alternate angles are equal) ……..(1)
⇒ ∠DAE = ∠BAE (Since, AE is the bisector of angle A) ………(2)
From equations (1) and (2), we get :
⇒ ∠AED = ∠DAE.
In △ DAE,
⇒ ∠AED = ∠DAE
⇒ AD = DE (In a triangle, sides opposite to equal angles are equal) …….(3)
From figure,
BE is the transversal.
⇒ ∠CEB = ∠EBA (Alternate angles are equal) ……..(4)
⇒ ∠CBE = ∠EBA (Since, BE is the bisector of angle B) ………(5)
From equations (4) and (5), we get :
⇒ ∠CEB = ∠CBE.
In △ CBE,
⇒ ∠CEB = ∠CBE
⇒ BC = CE (In a triangle, sides opposite to equal angles are equal) …….(6)
In parallelogram ABCD,
⇒ AB = CD (Opposite sides of parallelogram are equal)
⇒ AB = DE + EC
⇒ AB = AD + BC [From equation (3) and (6)]
⇒ AB = BC + BC (AD = BC, as opposite sides of parallelogram are equal)
⇒ AB = 2 BC.
Hence, proved that AB = 2 BC.
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In the alongside figure, ABCD is a parallelogram in which AP bisects angle A and BQ bisects angle B. Prove that :
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