Show that $\triangle ABK \cong \triangle ABL$. Show that $\triangle ABK \cong \triangle ABL$ where $D$ is the circumcenter,$K$ is the orthocenter and L lies on the circle.

 A: Here is a proof for an acute triangle $ABC$. I'll leave it up to you to extend it to triangles with an obtuse angle.

My diagram is like part of yours, with point $K$ the orthocenter of triangle $\triangle ABC$, but instead of point $L$ I have defined point $M$ on line $\overleftrightarrow{CK}$ such that $JK=JM$. Then by the two legs of a right triangle, $\triangle AJK \cong \triangle AJM$ and $\triangle BJK\cong\triangle BJM$. And those make $\triangle ABK\cong\triangle ABM$.
Let us call $\alpha=\angle ACK$ and $\beta=\angle BCK$. Then $\angle BAC$ is complementary to both $\alpha$ and $\angle ABK$, so $\angle ABK=\alpha$. So we also have $\angle ABM=\alpha$. Similarly, $\angle ABC$ is complementary to both $\beta$ and $\angle BAK$, so $\angle BAK=\beta$. So we also have $\angle BAM=\beta$.
Points $A,B,$ and $M$ make a triangle, so $\angle AMB$ is supplementary to the sum of $\alpha$ and $\beta$. That means $\angle AMB$ is supplementary to $\angle ACB$.
We now have supplementary, opposite angles in quadrilateral $ACBM$. That means quadrilateral $ACBM$ is a cyclic quadrilateral, meaning that points $A,C,B,$ and $M$ lie on the same circle. We see that the point $M$ that I defined is indeed the same as the point $L$ you defined as the intersection of line $\overleftrightarrow {CK}$ with the circumcircle of $\triangle ABC$.
We already know that $\triangle ABK\cong\triangle ABM$, so $\triangle ABK\cong\triangle ABL$, and we are done.
A: *

*We have side $AB$ in common.

*$\angle ABL =\angle ACL = 90 - \angle A=\angle ABK$

*$\angle BAL =\angle BCL = 90 - \angle B=\angle BAK$


Now, use SAS criteria, and we have $\triangle ABK \cong \triangle ABL$.
