Question

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

Answer 1

Step $1:$ Drawing the diagram:

A circle is drawn from the points $A,B$ and $C$ such that it intersects $CD$ at $E$.

$ABCD$ is a parallelogram

Step $2:$ Proving $\mathrm{AE}=\mathrm{AD}$:

From figure, we get,

$\angle$$\mathrm{ADE}$$+$$\angle$$\mathrm{ADC}=$$180°$…………(Linear pair)……$\left(1\right)$

As, $ABCE$ is a cyclic quadrilateral.

$\angle$$\mathrm{AED}$$+$$\angle$$\mathrm{ABC}=$$180°$…………(Sum of the Opposite angles is $180°$)………$\left(2\right)$

As, $ABCD$ is a parallelogram

$\angle$$\mathrm{ADC}=$$\angle$$\mathrm{ABC}$……………………(Opposite angles of parallelogram are equal)………..$\left(3\right)$

Using, $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$, we get,

$\angle$$\mathrm{ADE}$$+$$\angle$$\mathrm{ADC}=$$\angle$$\mathrm{AED}$$+$$\angle$$\mathrm{ADC}$,

Therefore, $\angle$$\mathrm{ADE}=$$\angle$$\mathrm{AED}$

Now, In $∆\mathrm{AED}$,

$\angle$$\mathrm{ADE}=$$\angle$$\mathrm{AED}$,

$\Rightarrow \mathrm{AE}=\mathrm{AD}$…………….(Sides opposite to equal angles are equal).

Hence, proved.