ABCD is a parallelogram. The circle through $ \mathrm{A}, \mathrm{B} $ and $ \mathrm{C} $ intersect $ \mathrm{CD} $ (produced if necessary) at $ \mathrm{E} $. Prove that $ \mathrm{AE}=\mathrm{AD} $.
Given:
ABCD is a parallelogram. The circle through \( \mathrm{A}, \mathrm{B} \) and \( \mathrm{C} \) intersect \( \mathrm{CD} \) (produced if necessary) at \( \mathrm{E} \).
To do:
We have to prove that \( \mathrm{AE}=\mathrm{AD} \).
Solution:
$ABCE$ is a cyclic quadrilateral.
We know that,
In a cyclic quadrilateral, the sum of the opposite angles is $180^o$.
This implies,
$\angle AEC+\angle CBA = 180^o$.........(i)
$\angle AEC+\angle AED = 180^o$.........(ii) (Linear pair)
From (i) and (ii), we get,
$\angle AED = ∠CBA$….....(iii)
$\angle ADE = \angle CBA$….........(iv) (Opposite sides of a parallelogram are equal)
From (iii) and (iv) we get,
$\angle AED = \angle ADE$
We know that,
Sides opposite to equal angles of a triangle are equal.
$AD$ and $AE$ are angles opposite to equal sides of a triangle.
Therefore,
$AD = AE$
Hence proved.
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