In the adjoining figure, M is the midpoint of side BC of a parallelogram


In the adjoining figure, M is the midpoint of side BC of a parallelogram ABCD such that ∠BAM = ∠DAM. Prove that AD = 2CD.


Given: parallelogram ABCD, M is the midpoint of side BC and ∠BAM = ∠DAM.

To prove: AD = 2CD

Since, $A D \| B C$ and $A M$ is the transversal.

So, $\angle D A M=\angle A M B \quad$ (Alternate interior angles)

But, $\angle D A M=\angle B A M$ (Given)

Therefore, $\angle A M B=\angle B A M$

$\Rightarrow A B=B M$

Now, $A B=C D \quad($ Opposite sides of a parallelogram are equal.)

$\Rightarrow 2 A B=2 C D$

$\Rightarrow(A B+A B)=2 C D$

$\Rightarrow B M+M C=2 C D \quad(A B=B M$ and $M C=B M)$

$\Rightarrow B C=2 C D$

$\therefore A D=2 C D \quad(A D=B C$, Opposite sides of a parallelogram are equal. $)$

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