Split the triangle in two right triangles whose lower or base side is $\frac{c}{2}$. Now these "new triangles" have angles, $60^\circ$, $30^\circ$ and $90^\circ$. As $\cos(30^\circ) = \frac{\sqrt{3}}{2} $, now you can easily check that $\frac{c}{2} = \frac{64 \cdot \sqrt{3}}{2} = 32 \cdot \sqrt{3}$. read more
The base, leg or altitude of an isosceles triangle can be found if you know the other two. A perpendicular bisector of the base forms an altitude of the triangle as shown on the right. This forms two congruent right triangles that can be solved using Pythagoras' Theorem as shown below. read more
The height of the isosceles triangle is one of the "legs" of the right triangles, and the other "leg" of each right triangle is half of the base (the right triangles are mirror images of each other, with the right angle formed between the base and height of the isosceles triangle). read more