Question

ABCD is a parlellogram.The circle through A,Band C intersect CD at E . Prove that AE = AD

Answer 1

GIVEN- A parallelogram ABCD
a circle through A,B,C intesects CD produced at E
TO PROVE- AE=AD
PROOF- angle ADE +angle ABC =180 ........1 (sum of opp. angle of a cyclic qurd.)
angle ADE +angle ADC=180 ...........2  (LINEAR PAIR)
angle ABC = angle ADC=180 .......3 (opp. angle of a parallogram are equal)
from 1 and 2
angle ADE +angle ADC = angle ABC = angle ADC
angle AED = angle ADE (using 3)
In triangle AED,
angle AED = angle ADE
AE = AD (equal sides have equal angles angles opp. to them)

HENCED  PROVED....

Answer 2

I think the proof is that    AE = BD,    AD = BE.  There is a mistake in the given question.

See the diagram.
ABCD is a parallelogram.  The circumcircle through A,B, & C intersects extended CD in E.

We know that in a circle, a chord inscribes the same angle at any point on the circumference on the same side.

Let  angle AED = x.    Let angle DAE = y  and  let angle ADB = z.

$\angle AED=\angle ABD=x\\As\ AB\ ||and\ DE,\\ AE\ (transversal)\ intersects\ both,\ \angle BAE=\angle AED=x.\\\\Transversal\ DB\ intersects\ parallel\ lines\ AB\ and\ DE,\\.\ \ \ \angle BDE=\angleABD=x.\\\\Chord\ DE\ makes\ \angle DBE\ =y\ and\ \angle DAE=\angle DBE=y\\\\Chord\ AB\ makes\ \angle ADB=z,\ and\ \angle AEB=\angle ADB=z\\Transversal\ BD\ intersects\ parallel\ lines\ AD\ and\ BC.\\Hence,\ \angle ADB=\angle DBC=z$


Compare the triangles  ADE and BDE.

  DE is the common side.  angle BDE = angle AED.   angle DAE = angle DBE.
  So the triangles are similar and as one side is common,  the triangles are congruent. 

       Hence,  AD = BE   and    AE = BD
             Since,  AD = BC,       AD = BE = BC.