Geometry on Isosceles Triangles

Question

As shown in the figure, $$\overline{BD}=\overline{DE}=\overline{AE}\\\overline{EH}=\overline{HF}=\overline{FI}=\overline{IG}=\overline{GC}$$ Where $\overline{EH}$ is the bisector of $\angle AEC$. What is the ratio of $\angle B$ and $\angle C$ ($\frac{\angle B}{\angle C}$) in degrees?

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I put all the statements on the figure, but it doesn't seem to give me much useful information.

I also tried to assume a specific angle (for example, $\angle B$) to be some unknown angle $x$, but even with a lot of isosceles triangles given, there are still a lot of unknown anges.

Answer 1

Let $\angle B = x$ and $\angle C = y$.

We have $\angle DEB = x$, and thus $\angle EDA = 2x$. $\angle DAE$ is $2x$ as well.

On the right we have $\angle GIC = y$. Hence $\angle IGF = 2y$, and so is $\angle IFG$.

Now $\angle FIH = 180^\circ - \angle FIG - \angle GIC = 180^\circ - (180^\circ - 2y - 2y) - y = 3y = \angle FHI$.

Similarly, $\angle HFE = \angle HEF = \angle HEA = 4y$.

Considering the angles at $E$, we have:

$$180^\circ = \angle BED + \angle DEA + \angle AEH + \angle HEF = x + (180^\circ - 4x)+4y+4y$$

Therefore $3x = 8y \leadsto \dfrac {\angle B}{\angle C} = \dfrac 83$.

Answer 2

Assume $\angle B$ and $\angle C$ to be $y$ and $x$ respectively.

By angle chasing, you can show that $\angle AEB=180-3y$ and $\angle HEC=4x$. Since $EH$ bisects $\angle AEC$, $\angle AEC=8x$

$\Rightarrow 3y=8x$

$\Rightarrow \frac{\angle B}{\angle C}=\frac{y}{x}=\boxed {\frac{8}{3}}$