Perpendiculars passing through diagonal intersection in a quadrilateral formed within a square

We drop perpendiculars from P which is the intersection of extensions of HN and FM on EF and HK. J and I are the foot of these perpendiculars respectively. We have:

$PF\bot BL , and, PJ\bot JF\Rightarrow \angle JPF=\angle BLF$

Also:

$\triangle LFB=\triangle PFG\Rightarrow PF=BL$

So right angled triangles PFJ and LFB are congruent due to ASA and we have:

$PJ=LF=\frac{EA}2$

Similarly triangles HPI and DHL are congruent and we have:

$PI=\frac {AK}2$

Also :

$AK=EA$

That is P has equal distance from EL and LK, so it is coincident on the intersection of diagonals of the square ELKA.

First note that the line $PD$ is the same length as $DL$, and so $\triangle PDL$ is an isosceles triangle with the same properties as your last question:

Prove that triangle CHL is isosceles in rectangle with specific construction

Let $X$ be the midpoint of $DH$, then $PX$ is perpendicular to $DH$, and $\angle XPL$ is $45^\circ$,and as $\angle PLD=135^\circ$, $PN$ is the same line as $PH$.

More generally, with liberal re-naming of points, consider not-necessarily-square rectangles $\square OACB$ and $\square ODFE$ with $D$ on line $OA$ and $E$ on line $OB$. We seek the conditions under which midpoint $M$ of $CF$ is such that $FA\perp MD$ and $FB\perp ME$.

Defining $a:=|OA|$, $b:=|OB|$, $d:=|OD|$, $e:=|OE|$, the reader can verify that the perpendicularity conditions yield the system $$\begin{align} (a - d)^2 &= e (b + e) \\[4pt] (b - e)^2 &= d (a + d) \end{align} \tag1$$ Solving for $d$ and $e$, we find $$(d,e) = \frac13(a,b)+\frac83k(-b,a) \qquad\text{or}\qquad (d,e) = \frac13(a,b)+\frac1{6k}(b,-a)\tag2$$ where $$k := \frac{a^2 - b^2}{a b + \sqrt{16 (a^2-b^2)^2 + a^2 b^2}} \tag3$$ For the square, $a=b$, we have that $k=0$, sending the second option in $(2)$ off to infinity, but reducing the first option to $\frac13(a,a)$, which is equivalent to OP's observation. $\square$

Rotate the rectangle $DHLE$ around the midpoint of $EL$, $90^o$ clockwise. I leave the details to you.

Proof:

Let $FC = HC = a$, so $DH = FB = 2a$.

Since $DE \parallel CF$ and $DE = CF = 2a$ with $\angle C = 90°$, quadrilateral $EFCD$ is a rectangle. Therefore $\angle DEL = 90°$, so $\angle AEL = 90°$.

Similarly, from rectangle $KBCH$ we have $\angle BKL = 90°$, so $\angle AKL = 90°$.

Also $\angle A = 90°$ and $AK = AE = 2a$.

Therefore $AKLE$ is a square with side length $2a$.

Extend segment $FM$ to meet diagonal $EK$ at point $P$.

Since $\angle HFE = \angle FEP = 45°$, we have $HF \parallel EP$ ...(1)

Extend line $FM$ to meet diagonal $EH$ of rectangle $ELHD$ at point $G$.

The rectangles $DELH$ and $LKBF$ are congruent, so their diagonals $EH$ and $LB$ are equal. Thus $\triangle EHL \cong \triangle LFB$, giving:

• $\angle LBF = \angle HEL = \varphi$ ...(2)
• $\angle BLF = 90° - \varphi$ ...(3)
• $\angle GLE = 90° - \varphi$ ...(4)

From (2) and (4): $\angle HEL + \angle GLE = 90°$, so $\angle EGL = 90°$.

Therefore $EH \perp MG$. Since $FM \perp BL$, we have $EH \parallel FP$ ...(5)

From (1) and (5), $EPFH$ is a parallelogram, so $EP = HF$ ...(6)

Triangles $\triangle EAK$ and $\triangle HFL$ are isosceles right triangles. By similarity: $$\frac{EK}{HF} = \frac{EA}{HL} = \frac{2a}{a} = 2$$

So $EK = 2HF$, and by (6), $EK = 2EP$.

Thus $P$ is the midpoint of diagonal $EK$, so $P$ is the center of square $AKLE$.

Since $P$ and $L$ lie on diagonal $AC$ of square $ABCD$, and $AC$ is an axis of symmetry, perpendiculars $HN$ and $FM$ are symmetric with respect to $AC$. Therefore $HN$ also passes through $P$. $\square$