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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- Sep 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 - Sep 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

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

Repeated use of isosoles triangles gives:

$\displaystyle m\angle LSD + 2m\angle LDS = 180$, and $\displaystyle m\angle ASD + 2m\angle SAD = 180$

$\displaystyle m\angle ASB + 2m\angle SAB = 180$, and $\displaystyle m\angle BSK + 2m\angle SBK = 180$

Thus:

$\displaystyle (m\angle ASD + m\angle LSD) + 2(m\angle LDS + m\angle SAD) = 360$

$\displaystyle (m\angle ASB + m\angle BSK) + 2(m\angle SAB + m\angle SBK) = 360$

From parallelogram m$\displaystyle \angle LDA = m\angle ABK$, so

$\displaystyle m\angle LDS + m\angle SDA = m\angle ABS + m\angle SBK$

But via isosoles, $\displaystyle m\angle SDA = m\angle SAD$, and $\displaystyle m\angle ABS = m\angle SAB$.

Thus

$\displaystyle m\angle ASD + m\angle LSD$

$\displaystyle = 360 - 2(m\angle LDS + m\angle SAD)$

$\displaystyle = 360 - 2(m\angle LDS + m\angle SDA)$

$\displaystyle = 360 - 2(m\angle ABS + m\angle SBK)$

$\displaystyle = 360 - 2(m\angle SAB + m\angle SBK)$

$\displaystyle = m\angle ASB + m\angle BSK$.

Now, using that $\displaystyle \overline{AN}$ is a diameter, get

$\displaystyle m\angle ASD + m\angle LSD+ m\angle NSL = 180$

$\displaystyle m\angle ASB + m\angle BSK+ m\angle NSK = 180$

Thus

$\displaystyle m\angle NSL = 180 - (m\angle ASD + m\angle LSD)$

$\displaystyle = 180 - (m\angle ASB + m\angle BSK)$

$\displaystyle = m\angle NSK$.

But from $\displaystyle m\angle NSL = m\angle NSK$ and Side-Angle-Side, get

$\displaystyle \triangle KSN \cong \triangle LSN$, and so $\displaystyle m\overline{NL} = m\overline{NK}$.