There is a parallelogram. Its vertexes A, D and B are tangent to the circle (triangle ADB is inscribed in the circle). Circle intersects parallelogram at points K and L. AN is diameter of the circle. Prove that |KN| = |KL|.

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- September 23rd 2012, 07:46 AMaleschioParallelogram and circle
There is a parallelogram. Its vertexes A, D and B are tangent to the circle (triangle ADB is inscribed in the circle). Circle intersects parallelogram at points K and L. AN is diameter of the circle. Prove that |KN| = |KL|.

Attachment 24909 - September 24th 2012, 02:16 PMTim28Re: Parallelogram and circle
SK,SN,and SL are radii and therefore equal. The two triangles are congruent. Congruent chords imply congruent arcs

- September 26th 2012, 06:24 AMHallsofIvyRe: Parallelogram and circle

You mean A, D, and B lie**on**the circle. Points cannot be 'tangent'.

Quote:

(triangle ADB is inscribed in the circle). Circle intersects parallelogram at points K and L. AN is diameter of the circle. Prove that |KN| = |KL|.

Quote: - September 26th 2012, 07:12 AMjohnsomeoneRe: Parallelogram and circle
It's all about the angles.

Repeated use of isosoles triangles gives:

, and

, and

Thus:

From parallelogram m , so

But via isosoles, , and .

Thus

.

Now, using that is a diameter, get

Thus

.

But from and Side-Angle-Side, get

, and so .