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.
 A: 
Let 


*

*$*sbh$ stands for triangles have same base and height. 

*$*c$ stands for triangles are congruent. 


We have


*

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

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

*$A_2 \stackrel{*c}= A_3$ 
A: 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.
A: 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 $A_4$? 
