A geometric problem where the result is obvious but the proof is hard... So M and N are the midpoints of the bases AB and CD of the isosceles trapezoid ABCD. The bisectors of angles CAD and ACB intersect with MN at points P and Q respectively. K and L are on AC so that the angles QLC and AKP are 90 degree.
Prove that the points K,L,M,N are on the same circle and if AD=c then find the radius of the circle.
I think that LNK=LMK=90 degree and LK=c and is the radius of the circle.
 A: Let $\angle ABC = a$ and $\angle ACD = b$. Other angles are marked in the diagram.

Then, $x = 0.5(180^0 – b – a) = … = 90^0 – 0.5(a + b); y = 0.5(a – b)$.
Also, $\angle PAM = b + y = … = 0.5(a + b)$
From $⊿APM, \angle APM = 90^0 – 0.5(a + b) = … = x$ [See above.]
∴ $APCQ$ is a cyclic quadrilateral. [Angles in the same segment equal]
By angles in the same segment, $\angle AQP = \angle ACP$ ….. (*)
Refer to the diagram below:-

$\angle ALQ = \angle AMQ = 90^0$ implies $ALMQ$ is a cyclic quadrilateral. [Angles in the same segment equal]
Similarly, $PCKN$ is also a cyclic quadrilateral.
$∴ h = \angle ACP$ [Ext. angle cyclic quad.]
$= \angle AQP$ [Proved in (*)]
$= k$ [Ext. angle cyclic quad.]
Thus, $LMKN$ is a cyclic quadrilateral [Angles in the same segment equal]
I believe $LK$ is the DIAMETER of that circle, not the radius.
A: The key observation is that $P$ and $Q$ lie on the circumcircle of $\square ABCD$. (Proof. Taking $P$ to be the midpoint of $\stackrel{\frown}{CD}$, we have that $\angle PAC$ and $\angle PAD$ subtend congruent arcs $\stackrel{\frown}{PC}$ and $\stackrel{\frown}{PD}$, and are therefore congruent; that is, $P$ is in fact the intersection of $\overleftrightarrow{MN}$ with the bisector of $\angle CAD$. Likewise for $Q$.)

Consider various right triangles with common vertex $X$, and hypotenuses 
$$a := |\overline{XA}| \qquad c := |\overline{XC}| \qquad p := |\overline{XP}|
 \qquad q:= |\overline{XQ}|$$
Clearly, all of these triangles are similar, with a common acute angle (the complement of the one at $X$) that we'll call $\theta$:
$$\angle XAM \cong \angle XCN \cong \angle XPK \cong \angle XQL \qquad= \theta$$
so that the opposing legs have lengths 
$$|\overline{XM}| = a \sin\theta \qquad |\overline{XN}| = c \sin\theta \qquad |\overline{XK}| = p \sin\theta \qquad |\overline{XL}| = q \sin\theta $$
We can conclude that $\square MLNK \sim \square AQCP$ (note the change in orientation!), with scale factor $\sin\theta$. Since the latter quadrilateral is cyclic, so is the former.
Now, by the Law of Sines applied to $\triangle ACD$, we have 
$$\frac{|\overline{AD}|}{\sin\theta} = \text{circumdiameter of }\triangle ACD = \text{circumdiameter of }\square AQCP$$
As $\sin\theta$ is the scale factor of our similar quadrilaterals, this implies 
$$|\overline{AD}| = \text{circumdiameter of }\square MLNK$$
