We are tasked with finding the ratio of $AC$ to $DE$ given that triangle $ABC$ is a right triangle with $\angle B = 90^\circ$, and triangle $BDE$ is equilateral. Additionally, the area of triangle $BDE$ is half of the area of triangle $ABC$.

Step 1: Notation and Setup

Let:

• $AB = c$,
• $BC = b$,
• $AC = a$.

The area of triangle $ABC$ is: $\text{Area of } \triangle ABC = \frac{1}{2} \cdot AB \cdot BC = \frac{1}{2} \cdot c \cdot b.$

Let $s$ denote the side length of the equilateral triangle $BDE$, i.e., $BD = BE = DE = s$. The area of an equilateral triangle with side length $s$ is given by: $\text{Area of } \triangle BDE = \frac{\sqrt{3}}{4} s^2.$

Step 2: Use the Given Condition on Areas

We are told that the area of triangle $BDE$ is half of the area of triangle $ABC$, i.e., $\frac{\sqrt{3}}{4} s^2 = \frac{1}{2} \cdot \frac{1}{2} \cdot c \cdot b.$ Simplifying this equation: $\frac{\sqrt{3}}{4} s^2 = \frac{1}{4} \cdot c \cdot b.$ Multiplying both sides by 4: $\sqrt{3} s^2 = c \cdot b.$ Thus, the side length $s$ of the equilateral triangle $BDE$ is: $s^2 = \frac{c \cdot b}{\sqrt{3}},$ $s = \sqrt{\frac{c \cdot b}{\sqrt{3}}}.$

Step 3: Ratio of $AC$ to $DE$

Now, we need to calculate the ratio $\frac{AC}{DE}$. Since $AC = a$ and $DE = s$, the ratio is: $\frac{AC}{DE} = \frac{a}{s}.$ We already know that $s = \sqrt{\frac{c \cdot b}{\sqrt{3}}}$, so the ratio becomes: $\frac{AC}{DE} = \frac{a}{\sqrt{\frac{c \cdot b}{\sqrt{3}}}}.$

Finally, this expression simplifies to give the desired ratio of $AC$ to $DE$.

Would you like further details on this simplification or have any questions?

Here are 5 related questions for further exploration:

1. How do the side lengths $AB$ and $BC$ affect the side length of $DE$?
2. What is the significance of the ratio $\frac{AC}{DE}$ in other triangle configurations?
3. How can we derive the formula for the area of an equilateral triangle?
4. What happens if the area of $BDE$ were a different fraction of $ABC$'s area?
5. How would the problem change if triangle $ABC$ were not a right triangle?

Tip: The relationship between area and side lengths often provides a path to discovering ratios in geometric problems.