Diagonal $A C$ of parallelogram $A B C D$ bisects $∠ A$ ( see the given figure). Show that $A B C D$ is a rhombus.

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Solution

Given: Diagonal $A C$ of parallelogram $A B C D$ bisects $∠ A .$
Then $∠ D A C = ∠ B A C \dots \dots (i)$ [ $A C$ bisects $∠ A$ ]
$∠ D A C = ∠ B C A \dots \dots (i i)$ [Alternate angles as $A B ∥ D C$ ]
and $∠ B A C = ∠ D C A \dots \dots (i i i)$ [Alternate angles as $A B ∥ D C$ ]
From the above three equations, we have
$∠ D A C = ∠ D C A = ∠ B C A = ∠ B C A$
In $Δ A D C$ , we have
$∠ D A C = ∠ D C A$ [from above]
$A D = C D$ [side opposite to equal angles]
Similarly, In $Δ A B C, A B = B C$
However, $A D = B C$ and $A B = C D$ (Opposite sides of a parallelogram)
$∴ A B = B C = C D = A D$
Hence, $A B C D$ is a rhombus.

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