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:

  1. $\triangle BDE$ is a 30-60-90 triangle so $BC= \frac{y\sqrt{3}}{2}$.
  2. $\triangle ABG$ is an isosceles triangle where $AB \cong BG$, so $AB=12$.
  3. $AC= \frac{y}{2}$
  4. $AC= AB+BC$, 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.

  1. I thought $BC= y\sqrt{3}$, not $\frac{y\sqrt{3}}{2}$. Why is it divided by $2$?
  2. How do we know $\triangle ABG$ is isosceles?
  3. I don't understand how $AC= \frac{y}{2}$.
  4. I can do step 4.

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  1. $BC$ 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.)

  2. 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.

  3. 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!

  • $\begingroup$ I have an addition question; how do we know AB is an altitude/angle bisector of angle FAG? I thought it might have something to do with the converse of the vertex angle bisector theorem, but wouldn't that mean we'd have to know it is an altitude? $\endgroup$ – Dana Mar 16 '16 at 2:18
  • $\begingroup$ Also, how do we prove that angle BCE is 90°/how do it we that from AB? $\endgroup$ – Dana Mar 16 '16 at 3:31
  • $\begingroup$ $B$ is the midpoint of the edge $FG$, and the large triangle is 45-45-90 (i.e., right isosceles) by assumption). Again, since everything is symmetric, it has to be 90*. (The two supplementary angles are equal.) $\endgroup$ – Xander Flood Mar 16 '16 at 4:24

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