Trivial by coordinate bash. Choose coordinates so that $A(-1,0)$, $B(1,0)$, $C(c,a)$, $D(d,a)$ - then the $y$-axis is the perpendicular bisector of $AB$. It suffices to compute the $x$-coordinates of $F$ and $H$ and to check that their sum (and hence their arithmetic mean) is $0$. This can be done in tons of ways, here is the fastest. Let $K$ and $L$ be the feet of the perpendiculars from $D$ and $F$ to $AB$. Then $\triangle ADK \cong \triangle FAL$ by hypotenuse and angles, so $AL = DK = a$, thus $L(-a-1,0)$ and the $x$-coordinate of $F$ is $-a-1$. Analogously the $x$-coordinate of $H$ is $a+1$ and we are done.

Complex numbers is very clean. Note $h-b=(b-c)i$, $f-a=(a-d)i$ so the midpoint of $FH$ is $\frac{a+b}{2}+\frac{(a+b)-(c+d)}{2}i. $ Orienting so that $\Im(a)=\Im(b)$ now solves the problem. This is because the perpendicular bisector of $AB$ is $\Re(z)=\frac{\Re(a)+\Re(b)}{2}$ and $(a+b)-(c+d)$ is purely real.

is there synthetic solution?

I swear this was on some Chinese middle school final exam... Construct point $X = AD \cap BC$, and $M$ the midpoint of $FG$. Reflect $A$ and $B$ over $M$ to get $A'$ and $B'$. We see that $FB'=BH=BC$ and $FA=AD$. Moreover, $\angle B'FA = \angle HFA + \angle BHF = 360^{\circ}-\angle FAB - \angle ABH = \angle BAD + \angle CBA - 180^{\circ} = \angle CXD$. Because $AB \parallel CD$, then $\frac{FB'}{FA} = \frac{BC}{AD} = \frac{XC}{XD}$. So $\triangle B'FA \sim \triangle CXD$. Finally, $\angle BAB' = 270^{\circ} - \angle FAB' - \angle DAB = 90^{\circ}$, so $BB' \perp AB$. Similarly, $AA' \perp AB$, meaning $ABA'B'$ is a rectangle with $M$ as its center. It is evident now that $X$ lies on the perpendicular bisector of $AB$.

There is no need of $E, G$. Translate $\triangle ADF$ of vector $\stackrel{\longrightarrow}{AB}$, $A$ goes to $B$, getting $D',F'$ so that $FF'\stackrel{\parallel}{=}DD'\stackrel{\parallel}{=}AB$. Well known, perpendicular from $B$ to $CD'$ goes through $P$, midpoint of $HF'$. Call $N$ midpoint of $AB$, construct rectangle $PBNM$, see that $H-M-F$ are collinear; with $MP\parallel FF'$ we are done. Best regards, sunken rock

yofro wrote: Complex numbers is very clean. Note $h-b=(b-c)i$, $f-a=(a-d)i$ Uhh correct me if I am wrong here. The idea is correct. The approach is definitely clean. But you have ( Probably because the solution is elementary and you rushed it ) a mistake in computing one of $f$ and $h$. (depends on orientation ). You can further check that yofro wrote: This is because the perpendicular bisector of $AB$ is $\Re(z)=\frac{\Re(a)+\Re(b)}{2}$ and $(a+b)-(c+d)$ is purely real. $(a+b)-(c+d)$ is never real since $AB \parallel CD$ Corrected it should be $(a+c)-(b+d)$

Helixglich wrote: yofro wrote: Complex numbers is very clean. Note $h-b=(b-c)i$, $f-a=(a-d)i$ Uhh correct me if I am wrong here. The idea is correct. The approach is definitely clean. But you have ( Probably because the solution is elementary and you rushed it ) a mistake in computing one of $f$ and $h$. (depends on orientation ). You can further check that yofro wrote: This is because the perpendicular bisector of $AB$ is $\Re(z)=\frac{\Re(a)+\Re(b)}{2}$ and $(a+b)-(c+d)$ is purely real. $(a+b)-(c+d)$ is never real since $AB \parallel CD$ Corrected it should be $(a+c)-(b+d)$ Sorry, one of $i$ should be replaced with $-i$, and I meant $(a+c)-(b+d)$. Everything else should check out

I the middle of the base AB, we choose the points M and Q on the base CD such that DM = QC = AI. Obviously, AIMD and BIQC are parallelograms, so IM = AD, IM || AD, and IQ = BC, IQ || BC. Outside the triangle IMQ, we construct the squares IMNP and IQRS. We consider the points U and V such that IPUS is a parallelogram, and V is the projection of I on CD. We have ∠PIS + ∠PIM + ∠MIQ + ∠QIS = 360°, and since IPUS is a parallelogram, we deduce ∠PIS + ∠MIQ = 180° = ∠IPU + ∠PIS, so ∠IPU = ∠MIQ. Since IP = IM and PU = IS = IQ, the triangles IPUS and MIQ are congruent (L.U.L.), so ∠PIU = ∠IMQ. The triangle IMV is right-angled at V, so ∠MIV = 90° - ∠IMV = 90° - ∠PIU. We obtain ∠MIV + ∠PIU + ∠MIP = 180°, so the points U, I, and V are collinear. Thus, UI is the median of the segment AB. Since IPUS is a parallelogram, the line UI passes through the midpoint J of the diagonal PS. From ∠DAF = ∠MIP = 90° and AD || IM, it follows that AF || IP and AF = IP, so the quadrilateral AIPF is a parallelogram. Thus, FP || AI and FP = AI. Similarly, we can show that SH || IB and SH = IB. Since I is the midpoint of the segment AB, we deduce that FP || HS and FP = SH, so FPHS is a parallelogram. Consequently, the diagonal FH passes through the midpoint J of the segment PS, so the lines UI and FH are concurrent

Ferum_2710 wrote: Let ∠AIP = ∠MIP = ∠MNP = ∠AFI = 90°. Let I be the midpoint of the base AB. We choose points M and Q on the base CD such that DM = QC = AI. Clearly, AIMN and BIQC are parallelograms, so IM = AD, IM ∥ AD, and IQ = BC, IQ ∥ BC. Outside the triangle IMQ, we construct the squares IMNP and IQRS. Let U and V be the points such that IPUS is a parallelogram, and V is the projection of I onto CD. We have ∠PIS + ∠PIM + ∠MIQ + ∠QIS = 360°. Since IPUS is a parallelogram, we deduce that ∠PIS + ∠MIQ = 180° = ∠IPU + ∠PIS, so ∠IPU = ∠MIQ. Since IP = IM and PU = IS = IQ, it follows that triangles IP U and MIQ are congruent (by L.U.L.), so ∠PIU = ∠IMQ. Triangle IMV is right-angled at V, so ∠MIV = 90° − ∠IMV = 90° − ∠PIU. We obtain ∠MIV + ∠PIU + ∠MIP = 180°, so the points U, I, and V are collinear. Thus, UI is the perpendicular bisector of segment AB. Since IPUS is a parallelogram, the line UI passes through the midpoint J of the diagonal PS. Since ∠DAF = ∠MIP = 90° and AD ∥ IM, it follows that AF ∥ IP, and since AF = IP, the quadrilateral AIPF is a parallelogram. Thus, FP ∥ AI and FP = AI. Analogously, it can be shown that SH ∥ IB and SH = IB. Since I is the midpoint of segment AB, we deduce that FP ∥ SH and FP = SH, so FPHS is a parallelogram. Consequently, the diagonal FH passes through the midpoint J of segment PS, so the lines UI and FH are concurrent. Please turn it into LaTeX