Question

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

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Solution

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 parallelogramStep $2:$ 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}$,$⇒\mathrm{AE}=\mathrm{AD}$…………….(Sides opposite to equal angles are equal).Hence, proved.

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