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.
$BC$ is equal to $CE\sqrt 3$, and $CE$ is half of $y$, so we have to divide by two. (The edges all have the same length, $CE$ is half an edge, and $y$ is a whole edge.)
if you fold the entire picture along the line $AB$, it'll line up with itself perfectly. In other words, angle $DFB$ and $EGB$ are equal. A triangle with two equal angls is isocoles.
In part 1, I mentioned why $CE$ is equal to $y/2$. Since $ACE$ is a 45-45-90 triangle (can you prove that?) it's two legs have the same length!