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

2
by bharti07dhiman

2015-02-23T13:43:18+05:30
GIVEN- A parallelogram ABCD
a circle through A,B,C intesects CD produced at E
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....

please mark it as the best
GIVEN- A parallelogram ABCD
a circle through A,B,C intesects CD produced at E
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
2015-02-23T14:14:26+05:30

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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.

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.