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

M mansi

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

To prove :   AE = AD

Proof :

$\angle$ADC = $\angle$3  , $\angle$ABC = $\angle$4, $\angle$ADE = $\angle$1  and $\angle$AED = $\angle$2

$\angle 3+\angle 1=180 \degree$.................1(linear pair)

$\angle 2+\angle 4=180 \degree$....................2(sum of opposite angles of cyclic quadrilateral)

$\angle$3 = $\angle$4      (oppsoite angles of parallelogram )

From 1 and 2,

$\angle$3+$\angle$1 = $\angle$2 + $\angle$

From 3,    $\angle$1 = $\angle$

From 4,    $\triangle$AQB,     $\angle$1 = $\angle$2

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

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