If the diagonal $A C$ of a parallelogram $A B C D$ bisects $∠$ $A$ , Show that $A B E D$ is a rhombus.

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Solution

Given $A C$ bisects $C$

$∠ 1 = ∠ 2$

T.P:- ABCD is rhombus

Sol:

$∠ 1 = ∠ 2$

$∠ 2 = ∠ 3$ (alternate opposite)

$∠ 1 = ∠ 4$

$\Rightarrow ∠ 3 = ∠ 4$

i.e., $∠ A C D = ∠ C A D$

$A D = C D$ (opposite sides of equal angle in a triangle are equal)

but $A B = C D$ & $A D = B C$

$\Rightarrow A B = B C = C D = D A$

Thus $A B C D$ is a rhombus.

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