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.

  • $\begingroup$ 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

A picture worths a thousand words


  • $*sbh$ 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 $A_4$?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.