Hello, leart369!

Label the inner right triangle like this:

Code:

A
*
| * c
b | *
| *
* - - - - - - - * B
C a

Note that for any triangle, its area is: .$\displaystyle A \;=\;\tfrac{1}{2}ab\sin C$

. . (One-half the product of two sides and the sine of the included angle)

The right triangle at the lower-left has sides $\displaystyle a$ and $\displaystyle b$.

. . Its area is: .$\displaystyle \boxed{A_1 \:=\:\tfrac{1}{2}ab}$

The triangle at the right has sides $\displaystyle a$ and $\displaystyle c$, and included angle $\displaystyle (180^o - B).$

. . Its area is: .$\displaystyle A_2 \:=\:\tfrac{1}{2}ac\sin(180^o-B) \:=\:\tfrac{1}{2}ac\sin B$

In the above diagram, we see that: .$\displaystyle \sin B \,=\,\tfrac{b}{c}$

. . Hence: .$\displaystyle A_2 \;=\;\tfrac{1}{2}ac\left(\tfrac{b}{c}\right) \quad\Rightarrow\quad \boxed{A_2 \:=\:\tfrac{1}{2}ab}$

The third triangle has sides $\displaystyle b$ and $\displaystyle c$, and included angle $\displaystyle (180^o - A).$

. . Its area is: .$\displaystyle A_3 \:=\:\tfrac{1}{2}bc\sin(180^o-A) \:=\:\tfrac{1}{2}bc\sin A$

In the above diagram, we see that: .$\displaystyle \sin A \,=\,\tfrac{a}{c}$

. . Hence: .$\displaystyle A_3 \;=\;\tfrac{1}{2}bc\left(\tfrac{a}{c}\right) \quad\Rightarrow\quad \boxed{A_3 \:=\:\tfrac{1}{2}ab}$

Fascinating! . . . All **four** triangles have the same area.