Q16 of 47 Page 10

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.

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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14 In figure, two circles touch internally at a point P. AB is a chord of the bigger circle touching the other circle at C. Prove that PC bisects the angle APB.[Hint: Draw a tangent at the point P. Joint CD, where D is the point of intersection of AP and the inner circle and prove that ∠ PBC = ∠ PCD.]
 15 Two circles intersect each other at two points A and B. At A, tangents AP and AQ to the two circles are drawn which intersect other circles at the points P and Q respectively. Prove that AB is the bisector of angle PBQ. 17 If PAB is a secant to a circle intersecting it at A and B, and PT is a tangent, then PA. PB = PT2.
(Prove using alternate segment theorem) 18 A circle touches the side BC of a triangle ABC at P and touches AB and AC when produced at Q and R respectively. Show that AQ = ½ (perimeter of triangle ABC).