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 $\mathrm{AE}=\mathrm{AD}$.

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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.

Updated on 10-Oct-2022 13:46:52