Answer :

Given: Diagonals of a parallelogram ABCD intersect in a point E.

To prove: Circumcircles of Δ ADE and Δ BCE touch each other at E.

Construction: Let

Proof:

∠ ADB = ∠ CBD (Alternate interior angles are equal)

∠ AEX = ∠ ADB (Alternate segment theorem)

But ∠ AEX = ∠ CEY (Vertically opposite angles)

∴ ∠ CBD = ∠ CEY

Therefore by converse of alternate segment theorem,

Hence the circumcircles of Δ ADE and Δ BCE touch each other at E.

To prove: Circumcircles of Δ ADE and Δ BCE touch each other at E.

Construction: Let

*l*be a tangent to first circle.Proof:

∠ ADB = ∠ CBD (Alternate interior angles are equal)

∠ AEX = ∠ ADB (Alternate segment theorem)

But ∠ AEX = ∠ CEY (Vertically opposite angles)

∴ ∠ CBD = ∠ CEY

Therefore by converse of alternate segment theorem,

*l*is a tangent to second circle with point of contact at E.Hence the circumcircles of Δ ADE and Δ BCE touch each other at E.

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