Question

if the angle between of Angle D and angle C of parallelogram most on AB at point. If, then prove that ad equal to 1 by 2 CD.​

Answer 1

Answer:

Given, AB∥CD, ∠D=2∠B, AD=b, CD=a

Now, AD=b ∴CM=b

and CD=a ∴AM=a

AB∥CD, thus, AMCD is a parallelogram.

∴∠D=∠AMC

∠AMC=2x

In △MCB

2x=∠MCB+∠x( ∵ Exterior angle is equal to sum of opposite interior angle)

=>∠MCB=x

=>∠MCB=∠B=x

∴MB=MC[As sides opposite to equal angles are equal]

Now, AB=AM+MB

=>AB=a+b