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

enter image description here

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.

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  • $\begingroup$ 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? $\endgroup$ – TZakrevskiy Sep 17 at 16:02
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A picture worths a thousand words

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

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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.

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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 $A_4$?

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