Prove that points A, B, K and L lie on a circle $c$ In an acute-angled triangle ABC with height CD, K and L are orthogonal projections through D respectively on AC and BC. Prove that points A, B, K and L lie on a circle $c$.
I tried to prove that triangles ADK and DLB are of the same form, but without any luck. 
 A: Hint: try to prove the following equalities and conclude from there:
$$\angle KAB=\angle CDK=\angle CLK$$
A: First notice that the $\angle CDK=\angle CBA$ because $\angle BKD$ and $\angle CDB$ are both $\pi/2$. For the same reason $\angle CDL=\angle CAB$. From this follows that $$\angle LDK=\angle CDK+\angle CDL=\angle CBA+\angle CAB$$ therefore $$\angle LDK+\angle ACB=\angle CBA+\angle CAB+\angle ACB=\pi.$$
This implies that the points $C,L,D,K$ lie in the same circle. Hence 
$$\angle DKL= \angle DCL.$$ Now notice that
$$\angle LKB+\angle CAB=\angle DKL+\pi/2+\angle CAB=\angle DCL+\pi/2+\angle CAB$$
and from the fact that $\angle DCB+\angle DBC=\pi/2$ we get 
$$\angle LKB+\angle CAB=\angle DCL+\angle DCB+\angle DBC+\angle CAB=\pi.$$
So $\angle LKB+\angle CAB=\pi$ and $A,L,K,B$ lie in the same circle.
A: $\begin{array}{} B⟂AB & BE∩KD={E} \\ AF⟂AB & AF∩DL={F} \\ \end{array}$
$ΔCDL∼ΔADF$
$\begin{array}{} \frac{CD}{DL}=\frac{DF}{AF} & ⇒ & AF=\frac{DL·DF}{CD} \end{array}$
$ΔCAD∼ΔBDE$
$\begin{array}{} \frac{CD}{AD}=\frac{BD}{BE} & ⇒ & BE=\frac{AD·BD}{CD} \end{array}$
$ΔADF∼ΔBDL$
$\begin{array}{} \frac{AD}{DF}=\frac{DL}{BD} & ⇒ & AD·BD=DL·DF \end{array}$
$\begin{array}{} AF=BE & ⇒ AE=BF & (ΔABE≅ΔABF)  \end{array}$
(angle in a semicircle): right triangles:
$\begin{array}{} ΔAKE, ΔABE & AE \text{(diameter)} & \overset{\Huge\frown}{AKBE}  \\ ΔABF, ΔBLF & BF\text{(diameter)} & \overset{\Huge\frown}{BLAF}  \end{array}$
$\begin{array}{} AE∩BF=M & \text{(Parallelogram Diagonals Theorem)}  \end{array}$
semicircles ($\overset{\LARGE\frown}{AKE}$ and $\overset{\LARGE\frown}{BLF})$ with congruent diameters ($AE = BF$) and common centers (midpoint $M$) are arcs contained in circle $M$ (circle $AKLB$)

