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  • ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

Q 6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that $A E= A D$.

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Given: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.

To prove:   AE = AD 

Proof:

$\angle \mathrm{ADC}=\angle 3, \angle \mathrm{ABC}=\angle 4, \angle \mathrm{ADE}=\angle 1$ and $\angle \mathrm{AED}=\angle 2$
$\angle 3+\angle 1=180^{\circ}$  (linear pair)--------(1)
$\angle 2+\angle 4=180^{\circ}$(sum of opposite angles of cyclic quadrilateral)-----------(2)
$\angle 3=\angle 4$   (opposite angles of parallelogram )
From 1 and 2,
$\angle 3+\angle 1=\angle 2+\angle 4$
From 3, $\angle 1=\angle 2$
From 4, $\triangle A Q B$
$\angle 1=\angle 2$

Therefore, $AE = AD$ (In an isosceles triangle, angles opposite to equal sides are equal)

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