2017 Japan MO P3

Just see the similarities for the triangles $AEF$ and $ABC$ $H$ for $AEF$ is like $K$ for $ABC$ $$\frac{AG}{GH}=\frac{AM}{MK}$$then in the end use the ratio and we are done

Let $a,b,c$ be complex numbers lying on unit circle such that they are vertices of the triangle. Then $$h=a+b+c,\ 2d=a+b+c-\frac{bc}{a},k=-a,\ m=\frac{a^2(b+c)}{a^2+bc}.$$Because quadrilaterals $BDHF, BCEF$ are cyclic we know that inversion centered at $A$ with radius $$\sqrt{AH\cdot AD}$$(or composition of inversion centered at $A$ with reflection wrt point $A$ if $\angle BAC>90^\circ$) transforms $D$ to $H$ and line $EF$ to circle $ABC$. Thus $G$ moves to second intersection of $AH$ with the circumcircle given by complex number $\frac{-bc}{a}$. Therefore $$(g-a)\left(-\frac{bc}{a}-a\right)=(h-a)(d-a)\iff g=\frac{(a-b)(a-c)(b+c)}{2(a^2+bc)}+a.$$We have $$g-m=\frac{(a-b)(a-c)(2a+b+c)}{2(a^2+bc)},\ h-k=2a+b+c.$$If $2a+b+c=0$ there holds $g=m,\ h=k$ and $GM\parallel HK$ is true. If $2a+b+c\neq0$ then $$\frac{g-m}{h-k}=\frac{(a-b)(a-c)}{2(a^2+bc)}$$and $$\overline{\left(\frac{g-m}{h-k}\right)}=\frac{(a-b)(a-c)}{2(a^2+bc)}=\frac{g-m}{h-k}\implies GM\parallel HK.\square$$#1716

$\frac{AK}{AM}=\frac{AH}{AG}$ Tugadi.