Prove the triangles $LBE$ and $LKD$ are congruent. I found this problem in a 2019 mathematics competition in Hunan, China. I am a Chinese high school student.

There is a triangle $ABC$ which is not equilateral. The point $H$ is the orthocenter of it. The circumscribed circle of the triangle has a tangent $BK$. The line $HK$ is perpendicular to $BK$. The midpoint of $AC$ is $L$. How to prove $LB$=$LK$?

I drew lines $AH$ and $CH$. Then I plotted a circle with the center $L$ and the diameter $AC$. Suppose the points $D$,$E$ are respectively the intersections of the circle $L$ and the lines $AH$,$CH$. Then I guess that the triangles $LBE$ and $LKD$ are congruent. But I can't prove it.

 A: This problem (original) is easy if you construct the diameter of the circumcircle $BM$.
Notice that $BM$ is the diameter, so $MA\perp AB$ and $MC\perp BC$. Also we have $AH\perp BC$ and $CH\perp AB$. So we have $AHCM$ is a parallelogram. Therefore, we have $M,L,H$ are collinear and $L$ is the midpoint of $MH$.
Now notice that $MB$ is the diameter, it pass the center of the circle, so $BM\perp BK$. Also we draw $LN\perp BK$ intersecting $BK$ at $N$. Since $BM, HK, LN$ are perpendicular to $BK$, we have those three segments are parallel. Furthermore, since we have $L$ is the midpoint of $HM$, we have $N$ is the midpoint of $BK$. So we have $LB=LK$.
For the second problem, your $D$ and $E$ are actually the height from $A$ to $BC$ and from $C$ to $AB$. So $LD=LE$. Furthermore we know that $DE\parallel BK$, since $LD=LE$ and $LB=LK$ we know that the angular bisector of $\angle DLE$ is the perpendicular line to $DE$, thus perpendicular to $BK$, thus parallel to the angular bisector of $\angle BLK$. So the angular bisector of $\angle DLE$ and that of $\angle BLK$ coincide, and thus we know that $\angle DLK=\angle ELB$. By SAS, we know that $\triangle LKD$ and $\triangle LBE$ are congruent.
A: Here is a picture proof for the one (remained) problem. We consider in it also the circumcenter $O$, and the mid points $9$ of $OH$, $B_1$ of $BH$, $K_1$ of $BK$. Then $9$ is the center of the nine-point circle $(9)$, containing $B_1,L$ as opposite points, and $HB_1OL$ is a parallelogram. (These are known properties. One way to see this, is as follows. The centroid $G$ is also on $OH$ so that $2=GH:GO=GB:GL=BH:OL$, so the segments $B_1H$ and $OL$ are parallel and of same length.)

It turns out that the points $K_1$, $B_1$, $9$, $L$ are collinear and on the side bisector of $BK$. Because of mid line properties, $K_1B_1\| KH\|BO\|B_1 9$, and $L\in B_19$. So $L$ is on the side bisector of $BK$, giving $LK=LB$.
$\square$
