$In the given figure, A B | | E C, A B = A C a n d A E b i s e c t s ∠ DAC.Prove that:$

$(i) ∠ E A C = ∠ A C B (i i) ABCE is a parallelogram.$

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Solution

Since $E D$ bisects $∠ D A C$ , therefore,
$\Rightarrow ∠ D A E = ∠ E A C$ ...(i)
Also, as $A B | | E C$ , therefore
$\Rightarrow ∠ B A C = ∠ E C A$ (Alternate interior angles) ...(ii)
Also, as $A B = A C$ , therefore, $∠ A B C = ∠ A C B$ ...(iii)
By Angle sum property, in $Δ A B C$
$\Rightarrow ∠ B A C = 180^{\circ} - 2 ∠ A C B$ [From (iii)] ...(iv)
Also, $B D$ is a straight line, therefore,
$\Rightarrow ∠ B A C = 180^{\circ} - 2 ∠ E A C$ [From (i)] ...(v)
From (iv) and (v), we get,
$\Rightarrow ∠ A C B = ∠ E A C$ [Hence proved the first part]
But they also form a pair of alternate interior angles.
Hence, $A E | | B C$
Therefore, $A B C E$ is a parallelogram as opposite pairs are parallel.

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