Question

ABCD is a parallelogram. the circle through A,B,C intersect CD at E . prove that AE=AD .
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Answer 1

To Prove: AE = AD.

Proof: In cyclic quadrilateral ABCE,

∠AED + ∠ABC = 180°    ...(1)

| ∵ Opposite angles of a cyclic quadrilateral are supplementary.

Also, ∠ADE + ∠ADC = 180°

| Linear Pair Axiom

But    ∠ADC = ∠ABC

| Opposite angles of a || gm

∴ ∠ADE + ∠ABC = 180°    ...(2)

From (1) and (2), we have

∠AED + ∠ABC = ∠ADE + ∠ABC

⇒    ∠AED = ∠ADE

∴ In triangle ADE,

AE = AD

| ∵ Sides opposite to equal angles of a triangle are equal. Proved.