There is a large right isosceles triangle with a hypotenuse length of $24$. Inside the triangle is an equilateral triangle with a vertex on the midpoint of the hypotenuse. If the length of each side of the equilateral triangle is $k(\sqrt{3}-1)$, find $k$.
I know that if $y$ is the length of one side of the equilateral triangle:
- $\triangle BDE$ is a 30-60-90 triangle so $BC= \frac{y\sqrt{3}}{2}$.
- $\triangle ABG$ is an isosceles triangle where $AB \cong BG$, so $AB=12$.
- $AC= \frac{y}{2}$
- $AC= AB+BC$, or $12= \frac{y\sqrt{3}}{2}+\frac{y}{2}$.
I'm just a bit confused as to reasoning behind the above/why it works.
- I thought $BC= y\sqrt{3}$, not $\frac{y\sqrt{3}}{2}$. Why is it divided by $2$?
- How do we know $\triangle ABG$ is isosceles?
- I don't understand how $AC= \frac{y}{2}$.
- I can do step 4.