Find The Ratio Of Areas

Let the side length of square $ABCD$ be $1$ . Then $AF = \dfrac 14$ and the area of $\triangle AEF$ , $[AEF] = \dfrac 12 \times \dfrac 14 \times 1 = \dfrac 18$ .

Since both $EF$ and $BD$ pass through the center of square $ABCD$ , it is clear to see that $G$ is the center of square $ABCD$ . Then the height of $\triangle DGE$ , $GI = \dfrac 12$ and its area $[DGE]=\dfrac 12 \times GI \times DE = \dfrac 12 \times \dfrac 12 \times \dfrac 34 = \dfrac 3{16}$ .

Let $HJ=a$ be perpendicular to $CD$ . Since $\triangle DHJ$ and $\triangle DBC$ are similar, $\dfrac {HJ}{DJ} = \dfrac {CD}{BC} = 1 \implies DJ=HJ = a$ . Similarly, $\triangle HEJ$ and $\triangle AED$ are similar and $EJ= \dfrac 34 a$ . We have $DJ+EJ = DE \implies a + \dfrac 34 a = \dfrac 34 \implies a = \dfrac 37$ . Then $[DHE]=\dfrac 12 \times \dfrac 34 \times \dfrac 37 = \dfrac 9{56}$ .

Now we have $\dfrac {[EGH]}{[AFGH]} = \dfrac {[DGE]-[DHE]}{[AEF]-[EGH]} = \dfrac {\frac 3{16}-\frac 9{56}}{\frac 18 - \frac 3{16}+\frac 9{56}} = \dfrac 3{11}$

Therefore $[EGH]:[AFGH] = \boxed{3:11}$ .

Let AB = a

(DGE) = (DE*Height)/2 = $3a^{2}$ /16

Triangle ABH ~ Triangle EDH

Therefore, AB/DE = Height JH/ Height IH = 4/3

So, IH = 3a/7

(EDH) = (DE*IH)/2 = $9a^{2}$ /56

(EGH) = (DGE) - (EHD) = $3a^{2}$ /112

(AFE) = (AF*AD)/2 = $a^{2}$ /8

(AFGH) = (AFE) - (EGH) = $11a^{2}$ /112

Therefore, (EGH)/(AFGH) = 3/11