+26367 Geometry Problem:
\(\text{Let the side of the square $= a$. $AD=DC=a$ } \\ \text{Let $DF = y$ } \\ \text{Let the area of triangle $[AED]=\dfrac{a^2}{2}$ } \\ \text{Let the area of triangle $[BCF]=\dfrac{(a-y)a}{2}=S_2+S_4+S_7$ } \\ \text{Let the area of triangle $[ADF]=\dfrac{ay}{2}=S_5+S_8$ } \)
\(\begin{array}{|rcll|} \hline \dfrac{(a-y)a}{2} &=& S_2+S_4+S_7 \\ \mathbf{S_4} &=& \dfrac{(a-y)a}{2} - S_2 - S_7 \\ \hline \dfrac{ay}{2} &=& S_5+S_8 \\ \mathbf{S_5} &=& \dfrac{ay}{2} - S_8 \\ \hline \mathbf{S_3} &=& \mathbf{[AED] - S_4 -S_5} \\ S_3 &=& \dfrac{a^2}{2} -\left( \dfrac{(a-y)a}{2} - S_2 - S_7 \right) - \left(\dfrac{ay}{2} - S_8 \right) \\ S_3 &=& \dfrac{a^2}{2}-\dfrac{a^2}{2} +\dfrac{ay}{2} + S_2 + S_7 - \dfrac{ay}{2} + S_8 \\ S_3 &=& S_2 + S_7 + S_8 \\ S_3 &=& 2 + 8 + 2 \\ \mathbf{S_3} &=& \mathbf{12} \\ \hline \end{array}\)
