In a parallelogram ABCD the bisector of ∟A also bisects BC at P then how do you prove that AD = 2 AB?
Given AP is the angle bisector of ∠A and also bisects BC, so let ∠PAB=∠PAE=x and PB=PC=y Draw PE||BA =>∠PAB=∠APE=x {interior alternate angles} so, in ΔPAE, ∠PAE=∠APE=x according to the theorem that sides opposite to equal angles in a triangle are equal. =>AE=PE=y since, PE=AB=CD=y =>AD=2AB Hence Proved
M CUBE: MatheMatics by Maheshwari 