# Parallelogram and circle

• Sep 23rd 2012, 07:46 AM
aleschio
Parallelogram 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 PM
Tim28
Re: 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 AM
HallsofIvy
Re: Parallelogram and circle
Quote:

Originally Posted by aleschio
There is a parallelogram. Its vertexes A, D and B are tangent to the 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 AM
johnsomeone
Re: Parallelogram and circle

Repeated use of isosoles triangles gives:

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

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

Thus:

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

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

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

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

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

Thus

$m\angle ASD + m\angle LSD$

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

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

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

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

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

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

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

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

Thus

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

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

$= m\angle NSK$.

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

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