Geometric Construction: Restore the triangle with the only points marked are orthocentre and one vertex Given obtuse triangle $BAC$ with $\angle A$ obtuse. Let $H$ be the orthocentre. Let $M$ be the midpoint of $BC.$ Define $X$ such that $MB=MC=MX, D\in AH.$ Let $l$ be line $XM.$ Define $k$ as line $XB$ and $t$ as line $XC.$ Now everything is deleted except points $H,A,X$ and $l,k,t.$
Can we reconstruct $ABC$ triangle?
For this problem, I defined the line perpendicular to $AH$ at $H.$ Let the line intersect $k,t$ at $F, G.$ Note that $GDF\sim CDB.$
Also it's well known that $DA,DM$ is isogonal.Therefore $DA$ is the symmedian of triangle $DCB.$
So we can use harmonic conjugates,etc. Consider $DD$ wrt $DCB$ then $DD\perp DM,$ as $M$ is centre. I also tried to use this famous geometry lemma, which states

if $ABC$ a trinagle and $H$ is the orthocentre, $M$ is the midpoint of $BC$ then $M-G-H$ collinear where $G=(AH)\cap (ABC).$ We have $(T,D;B,C)=-1.$ And $H$ is orthocentre of $ATM.$

Also poles and polars maybe?
Any ideas?
The triangle before deleting:

After deleting:

The lemma diagram:

 A: 
Here is the construction you need to recover the deleted $\triangle ABC$ using the lines $k$, $l$, and $t$ and points $A$, $H$, and $X$.
We start our construction by joining $A$ and $H$ to reconstruct the altitude $AH$ of $\triangle ABC$. Now, draw the lines $KL$ and $DE$ perpendicular to $AH$ through $H$ and $A$ respectively, so that they meet lines $k$ at $L$ and $D$, and lines $t$ at $K$ and $E$. When you draw the two lines $LE$ and $KD$ as shown in the diagram, they intersect each other at a point $N$, which is located on the given line $l$. The midpoint of the line segment of $XN$ is the midpoint $M$ of the side $BC$ of the deleted $\triangle ABC$. We finish the construction by drawing a line through $M$ perpendicular to $AH$ to cut the lines $k$ and $t$ at $B$ and $C$ respectively.
We kindly invite OP to provide a proof of this construction.
$\underline{\color{red}{Note}}:$
The restoration of $\triangle ABC$ is possible even when no information about the line $l$ is available because the most vital bit of information, i.e. the whereabouts of $N$, can be obtained by letting the two lines $KD$ and $LE$ intersect each other.
