Orthocenter in obtuse triangle from square and equilateral triangle configuration

My Proof (Synthetic)

Let $T$ be the intersection of $KM$ with $EZ$.

Since $E$ and $Z$ are midpoints of opposite sides, $ABZE$ is a rectangle, so $KB \parallel EZ$. Thus $\angle KZE = \angle KZL - \angle EZL = 90° - 60° = 30°$ and $\angle BKZ = 30°$. In right triangle $KBZ$ we have $BZ = \frac{KZ}{2}$, so $2 \cdot BZ = BC = KZ = EZ = ZL$. This makes triangle $KZL$ isosceles right-angled with $\angle ZLK = \angle ZKL = 45°$.

The perpendicular $ZH$ from $Z$ to $KL$ is also a median (since $KZ = ZL$), so $KH = LH$. By SAS, $\triangle MHK \cong \triangle MHL$ (with $MH$ common, $\angle MHK = \angle MHL = 90°$, and $KH = LH$), giving $\angle MKH = \angle MLH$. Since $\angle KLE = \angle ELZ - \angle KLZ = 60° - 45° = 15°$, we have $\angle MKH = 15°$.

In triangle $KTZ$: $$\angle KTZ = 180° - \angle TKP - \angle PKZ - \angle TZK = 180° - 15° - 45° - 30° = 90°$$

Therefore $KM \perp EZ$. Since $P \in EZ$, this gives $MK \perp PZ$. Combined with $MZ \perp KP$ (from construction: $ZM \perp KL$ and $P \in KL$), the two altitudes of triangle $KPZ$ from vertices $K$ and $Z$ intersect at $M$.

Therefore, $M$ is the orthocenter of triangle $KPZ$. Note that triangle $KPZ$ is obtuse, so the orthocenter lies outside the triangle.

In the picture we have:

$KZ\bot ZL$

$ZM\bot KL$

$\Rightarrow \angle KZM=\angle LZM$

This means in triangle KZL , ZH is the bisector of angle KZL and is also perpendicular to the base K. This means that triangle KZL is isosceles and we have:

$KZ=ZL$

M is on the perpendicular bisector of KL so triangle MKL is isosceles and we have:

$MK=ML\Rightarrow \angle MKL=\angle MLK\space\space\space\space\space\space(1)$

Also:

$LE=ZE$

$ZM= LP$

So triangles ZME and LEP have two sides equal with common angle PEM, so they are congruent and we have:

$\angle EZM=\angle ELP\space\space\space\space\space\space(2)$

(1) and (2) results in:

$\angle MKH=\angle EZM$

and we have:

$ZM\bot KP\Rightarrow KM\bot EZ$

This means that M is the orthocenter of triangle KPZ.

$PZLM$ is an isosceles trapezoid, therefore $PM||ZL$.

Similarly, $ZHTK$ is an isosceles trapezoid, therefore $HT||ZK$.

As $ZL\perp ZK$,$PM\perp HT$, so we are done.

A trigonometric method is to note that $|ZK|=|ZE|$ as so $\triangle EZK$ is isosoceles, therefore $\angle KEZ=75^\circ$.

Therefore $\angle EKT=15^\circ=\angle PKT$, therefore $\triangle EMP$ is equilateral, so $\angle KMP=30^\circ$ as required.

Other points are $TPHM$ is cyclic, and as $|TZ|=\frac{\sqrt3}{2}$, $|ET|=1-\frac{\sqrt3}{2}$ and $|ZP|=\sqrt3-1$.