Let $CD,BE$ meet $(ABC)$ at $T,S$ respectively. Let $TS$ meet $AB,AC$ at $F',G'$ respectively. Note that $\angle F'BD=\angle SBC=\angle F'TD$, so $TF'DB$ is concyclic. This means that $\angle BF'D=\angle BTC=\angle BAC$, so $F'=F$. Similarly, $G'=G$. Now $\angle BFG=180^\circ-\angle TFB=180^\circ-\angle TDB=\angle BDC$, as desired.

Complex bash with $\triangle ABC$ inscribed in the unit circle, and let lines $BD,CE,BE,CD$ intersect the unit circle again at $X,Y,Z,W$ respectively, so that $$|a|=|b|=|c|=|x|=|y|=1$$$$z = \frac{ac}x$$$$w = \frac{ab}y$$$$d = \frac{bx(c+w) - cw(b+x)}{bx-cw} = \frac{abc+acx-abx-cxy}{ac-xy}$$$$e = \frac{cy(b+z) - bz(c+y)}{cy-bz} = \frac{abc+aby-acy-bxy}{ab-xy}$$Now let the line through $D$ parallel to $AC$ intersect the unit circle at $U,V$. Then $$uv = ac$$and since $d = u + v - \overline{d}uv$, we have $$u + v = d + ac\overline{d} = \frac{b(abc+acx-abx-cxy) + ac(ab+cy-by-xy)}{b(ac-xy)} = \frac{a^2bc+ab^2c+abcx+ac^2y-ab^2x-abcy-acxy-bcxy}{b(ac-xy)}$$Then \begin{align*} f &= \frac{ab(u+v) - uv(a+b)}{ab-uv} \\ &= \frac{\frac{a(a^2bc+ab^2c+abcx+ac^2y-ab^2x-abcy-acxy-bcxy)}{ac-xy} - ac(a+b)}{ab-ac} \\ &= \frac{(a^2bc+ab^2c+abcx+ac^2y-ab^2x-abcy-acxy-bcxy) - c(a+b)(ac-xy)}{(b-c)(ac-xy)} \\ &= \frac{a^2bc-a^2c^2+ab^2c-abc^2+abcx+ac^2y-ab^2x-abcy}{(b-c)(ac-xy)} \\ &= \frac{a^2c+abc-abx-acy}{ac-xy} \end{align*}Similarly, $$g = \frac{a^2b+abc-abx-acy}{ab-xy}$$Then we find the vectors \begin{align*} b-d &= \frac{x(a-y)(b-c)}{ac-xy} \\ c-d &= -\frac{a(b-c)(c-x)}{ac-xy} \\ b-f &= -\frac{(a-y)(ac-bx)}{ac-xy} \\ g-f &= -\frac{a^2(b-c)(b-y)(c-x)}{(ab-xy)(ac-xy)} \end{align*}Then $$\frac{\left(\frac{b-f}{g-f}\right)}{\left(\frac{b-d}{c-d}\right)} = -\frac{(ac-bx)(ab-xy)}{ax(b-c)(b-y)}$$This is equal to its conjugate and thus real. $\blacksquare$