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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)

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

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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.