Question

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

Answer 1

Given:$ABCD$ is a parallelogram
To prove: $AE=AD$
Construction: Draw a circle which passes through $ABC$ and intersect $CD$ (or $CD$ produced) at $E$.
Proof: Case I
$\angle AED+\angle ABC={180}^{\circ }$ [opposite angles of cyclic quadrilateral].....(i)
$\angle ABC=\angle ADC$ [opposite angles of parallelogram are equal].........(ii)
$\angle ADC+\angle ADE={180}^{\circ }$ ($\because$ Linear Pair) .......(iii)
Substituting (ii) in (iii) we get
$\angle ABC+\angle ADE={180}^{\circ }$ ........(iv)
From (i) and (iv)
$\angle AED+\angle ABC=\angle ABC+\angle ADE$
$\Rightarrow \angle AED=\angle ADE$
$\Rightarrow AD=AE$ [Sides opposite to equal angles are equal]

Similarly we can prove Case II.