Overkill

Bruh....

Brilliant problem. Note $K$ is the center of the spiral similarity sending $FB$ to $CG$, whence it sends the squares $AEFB$ to $DACG$. Let $O_1$ and $O_2$ be the centers of these two squares respectively, whence it also sends $O_1$ to $O_2$. Set $N$ to be the midpoint of $BC$ and let $M$ be the midpoint of $O_1O_2$. Then by the mean geometry theorem we have that $MN \perp O_1O_2$ and $MN = MO_1 = MO_2$. Now, note that \[ \measuredangle FKC = \measuredangle FKI + \measuredangle IKC = \measuredangle FBI + \measuredangle IGC = \measuredangle (BF, GC) = \measuredangle BAC, \]so particularly from $\frac{FK}{KC} = \frac{AB}{DG} = \frac{AB}{AC}$ we obtain $\triangle FKC \sim \triangle BAC$, and similarly $\triangle KO_1O_2 \sim \triangle ABC$. Note these are positive similarities. Length chasing obtains that $A$ lies on the $K$-apollonian circle of $\triangle KO_1O_2$, hence so does $H$ as it is the reflection of $A$ across $O_1O_2$. Moreover we can compute $\measuredangle O_1AO_2 = \measuredangle O_1KO_2 + 90^\circ$, so $(O_1AO_2)$ is orthogonal to $(KO_1O-2)$. We look at the image of $\triangle KO_1O_2$ under an inversion at $A$. Note that $K$ lies on a line orthogonal to some circle, the image of $KO_1O_2$, and then $O_1$ and $O_2$ lie on a line orthogonal to the same circle, but also perpendicular to the line through $K$. So the image of $KO_1O_2$ is a right isosceles triangle, with $O_1K = O_2K$. Particularly we obtain $\measuredangle NO_1H = \measuredangle AKO_2$, and $\measuredangle NO_2H = \measuredangle AKO_1$. Now a similar argument yields that the image of $\triangle NO_1O_2$ after an inversion at $H$ is a triangle directly similar to $KO_1O_2$; indeed we see this by noting the image of $HO_1O_2$ is directly similar to $AO_1O_2$, and then the angle conditions force the location of the image of $N$, which forces a similarity just by comparing to the inversion at $A$. Perform a $\sqrt{HO_1 \cdot HO_2}$ inversion at $H$ along with a reflection across the angle bisector. Then we note that $N$ must go to the reflection of $K$ across $O_1O_2$, say $K'$. This means that $HK'$ and $HN$ are isogonal in $\triangle O_1HO_2$. It suffices now to show that $KH$ and $K'H$ are isogonal in this triangle too, but this is not hard as we note the angle bisector of $O_1HO_2$ intersects $O_1O_2$ at a point on $(KK'AH)$, precisely an arc midpoint of $KK'$, which establishes the result.

Here's a toy complex bash. $a, b, c$. $e = a + (a-b)i$ $f = b + (a-b)i$ $d = a + (c-a)i$ $g = c + (c-a)i$ $k = (fg - bc)/(f + g - b - c) = a + i \cdot \frac{(a-b)(a-c)}{b-c}$. $m = \frac{b+c}{2}$. $ h = \frac{(1 + i) (a^2b - i ab^2 - a^2c + ib^2c - i ac^2 + ibc^2)}{2(ab - iac - (1 - i)bc)}$. It remains to show that $2a + i \cdot \frac{2(a-b)(a-c)}{b-c}$, $b+c$, $\frac{(1 + i) (a^2b - i ab^2 - a^2c + ib^2c - i ac^2 + ibc^2)}{ab - iac - (1 - i)bc}$ are collinear. Clearing denominators, this becomes $(2a(b-c) + i \cdot 2(a-b)(a-c)) \cdot (ab - iac - (1 - i)bc)$, $(b + c)(b-c)(ab - iac - (1 - i)bc)$, $(1 + i)(b-c) (a^2b - i ab^2 - a^2c + ib^2c - i ac^2 + ibc^2)$ are collinear. We then want to show that \[ \frac{(b-c)((b + c)(ab - iac - (1 - i)bc) - (1 + i)(a^2b - i ab^2 - a^2c + ib^2c - i ac^2 + ibc^2))}{(b^2 - c^2 - (2a(b-c) + i \cdot 2(a-b)(a-c)))(ab - iac - (1 - i)bc)} = \frac{a (b - c)^2 ((-1 - i) a + i b + c)}{((-1 + i) a + b - i c)^2(ab - iac - (1 - i)bc)} \]is a real but it's self conjugate so the result follows.

Let $O_1$ and $O_2$ be the centers of $AEFB$ and $ADGC$ and let $M$ be the midpoint of $BC$ and $AA'$. Claim: $KO_1O_2$ and $ABC$ are similar By construction, $K$ is the center of the spiral symmetry $GC\rightarrow BF$ and thus is also the center of the spiral symmetry $ADGC \rightarrow EABF$ and $O_2\rightarrow O_1$. As the squares have an angle of $\angle BAC$ between them and have sides of $AC$ and $AB$ we can express the spiral symmetry as rotation by $\angle BAC$ about $K$ followed by a homothety with factor $AB/AC$ about $K$. Notice that the consequence of this is that triangles $ABC$ and $KO_1O_2$ are similar. Claim: $MO_1O_2$ is a right isosceles triangle Doubling about $A$ we show that $A'FG$ is a right isosceles triangle. Notice $\angle A'BF=\angle A'CG=90^{\circ}-\angle BAC$, $GC=A'B=AC$, and $FB=A'C=AB$. Thus $A'G=A'F$ and $\angle GA'F=90^{\circ}$. As $A'G\parallel BD$ and $A'F\parallel CE$ we have that $\angle BHC=90^{\circ}$. Claim: Triangles $MBO_1$ and $MO_1K$ are similar First, $\angle MBO_1=\angle MO_1K=45^{\circ}+\angle ABC$. Then $$\frac{O_1K}{MO_1}=\sqrt{2}\cdot \frac{O_1K}{O_1O_2}=\sqrt{2}\cdot \frac{AB}{BC}=\frac{O_1B}{BM}$$ To finish $$\angle KMB=2\angle O_1MB=2\angle ECB=\angle HMB$$

Attachments:

YaoAOPS wrote: Here's a toy complex bash. $a, b, c$. $e = a + (a-b)i$ $f = b + (a-b)i$ $d = a + (c-a)i$ $g = c + (c-a)i$ $k = (fg - bc)/(f + g - b - c) = a + i \cdot \frac{(a-b)(a-c)}{b-c}$. $m = \frac{b+c}{2}$. $ h = \frac{(1 + i) (a^2b - i ab^2 - a^2c + ib^2c - i ac^2 + ibc^2)}{2(ab - iac - (1 - i)bc)}$. It remains to show that $2a + i \cdot \frac{2(a-b)(a-c)}{b-c}$, $b+c$, $\frac{(1 + i) (a^2b - i ab^2 - a^2c + ib^2c - i ac^2 + ibc^2)}{ab - iac - (1 - i)bc}$ are collinear. Clearing denominators, this becomes $(2a(b-c) + i \cdot 2(a-b)(a-c)) \cdot (ab - iac - (1 - i)bc)$, $(b + c)(b-c)(ab - iac - (1 - i)bc)$, $(1 + i)(b-c) (a^2b - i ab^2 - a^2c + ib^2c - i ac^2 + ibc^2)$ are collinear. We then want to show that \[ \frac{(b-c)((b + c)(ab - iac - (1 - i)bc) - (1 + i)(a^2b - i ab^2 - a^2c + ib^2c - i ac^2 + ibc^2))}{(b^2 - c^2 - (2a(b-c) + i \cdot 2(a-b)(a-c)))(ab - iac - (1 - i)bc)} = \frac{a (b - c)^2 ((-1 - i) a + i b + c)}{((-1 + i) a + b - i c)^2(ab - iac - (1 - i)bc)} \]is a real but it's self conjugate so the result follows. Actually your work can be simplified a lot by letting $b=-1,c=1,$ here the midpoint is $0,$ everything is more friendly.