Question

In a parallelogram ABCD, the bisector of A also bisects BC at X. Prove that AD = 2 AB

Answer 1

Step-by-step explanation:

Given, ABCD is a parallelogram.

∴ AD||BC (Opposite sides of the parallelogram are parallel)

Now, AD||BC and AX is the transversal ,

∴ ∠2 = ∠3 (Alternate angles) ............(1)

and ∠1 = ∠2 (AX is the bisector of ∠A) ................(2)

From (1) and (2), we obtain

∠1 = ∠3

Now, in ΔABX,

∠1 = ∠3

⇒ AB = BX ( If two angles of a triangle are equal, then sides opposite to them are equal)

⇒ 2AB = 2BX = BX + BX = BX + XC ( X is the mid point of BC)

⇒ 2AB = BC

⇒ 2AB = AD (Opposite sides of a parallelogram are equal)

∴ AD = 2AB.