Triangle area proof

Squares are constructed above the overhangs and the transom. Connecting the outer vertices of adjacent squares creates three triangles. Prove that their areas are the same.

Final Answer:

x = 1

Step-by-step explanation:

$$\begin{gathered} a,b,c \\ {c}^2=a^2+b^2 \\ \alpha +\beta =90{}^{\circ } \\ \sin \beta =b:c \\ \sin \alpha =a:c \\ {S}_1=\frac{a\cdot b}{2} \\ {S}_2=\frac{c\cdot b\cdot \sin \alpha }{2} \\ {S}_3=\frac{c\cdot a\cdot \sin \beta }{2} \\ {S}_1=\frac{a\cdot b}{2} \\ {S}_2=\frac{c\cdot b\cdot a/c}{2} \\ {S}_3=\frac{c\cdot a\cdot b/c}{2} \\ {S}_1=\frac{a\cdot b}{2} \\ {S}_2=\frac{a\cdot b}{2} \\ {S}_3=\frac{a\cdot b}{2} \\ {S}_1=S_2 \\ x=1 \end{gathered}$$

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