Question

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

Answer 1

Data: $ABCD$ is a parallelogram. The circle through $A,B$ and $C$ intersect $CD$ at $E$ is joined.
To prove: $AE=AD$
Proof: $\angle AEC+\angle AED={180}^o$........(i)(linear pair)
$ABCE$ is a cyclic quadrilateral.
$\therefore \angle ABC+\angle AEC={180}^o$.........(ii)(Opposite angles)
Comparing (i) and (ii)
$\angle AEC=\angle AED=\angle ABC=\angle AEC$
$\angle AED=\angle ABD$...........(iii)
But, $\angle ABC=\angle ADE$ (opposite angles of quadrilateral)
Substituting in equation (iii),
$\angle AED=\angle ADE$
$\therefore AE=AD$.