# Prove that the area of the trangles are equal.

Prove that the area of all the traingles in the figure below are equal.

I tried using geogebra to determine an arbitrary values of $$a$$ , $$b$$, and $$c$$. I found out that the triangles have equal measure of area.

• Hint: what is the angle between sides "with two stripes" and "with three stripes" in your triangle $A_1$? How is it related to angles in the triangle $A_3$? How do we find are given two sides and an angle between them? – TZakrevskiy Sep 17 at 16:02

Let

• $$*sbh$$ stands for triangles have same base and height.
• $$*c$$ stands for triangles are congruent.

We have

1. $$A_1 \stackrel{*sbh}{=} B_1 \stackrel{*c}{=} B_3 \stackrel{*c}{=} A_3$$,

2. $$A_3 \stackrel{*c}{=} B_3 \stackrel{*c}{=} B_4 \stackrel{*sbh}= A_4$$,

3. $$A_2 \stackrel{*c}= A_3$$

It is useful to know that the area of a triangle can be calculated by $$\frac12ab\sin\theta$$ where $$a$$ and $$b$$ are two side lengths of the triangle and $$\theta$$ is the angle between those two side lengths.

Let $$\alpha$$ be the angle in $$A_3$$ formed by the side lengths $$b$$ and $$c$$. Then, the angle formed by the side lengths $$b$$ and $$c$$ in $$A_4$$ is $$\pi-\alpha$$. Since $$\sin\alpha=\sin(\pi-\alpha)$$, $$A_3$$ and $$A_4$$ have the same area.

This can be applied to the other triangles in your diagram.

Area of $$A_2$$ and $$A_3$$ is $$ab/2$$.

For $$A_1$$: draw a parallel to $$b$$ through the upper vertex of the square on $$c$$. The distance from this parallel to the side at bottom of $$A_1$$ is $$b$$ (Pythagoren theorem).

Can you do the same for $$A_4$$?