Q. 16

The diagonals of a parallelogram ABCD intersect in a point E. Show that the circumcircles of Δ ADE and Δ BCE touch each other at E.

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 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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