A Lonely Point Outside the Square Becomes an Identity!

$\triangle EFC \Rightarrow EF^2 + FC^2 = EC^2 \space \cdots (1)$ .

$\triangle AEG \Rightarrow AG^2 + EG^2 = AE^2 \space \cdots (2)$ .

$(1) + (2) \Rightarrow AE^2 + EC^2 = AG^2 + EG^2 + EF^2 + FC^2 \space \cdots (3)$ .

$\triangle BFE \Rightarrow BF^2 + EF^2 = BE^2 \space \cdots (4)$ .

$\triangle DEG \Rightarrow EG^2 + DG^2 = DE^2 \space \cdots (5)$ .

$(4) + (5) \Rightarrow DE^2 + BE^2 = EG^2 + DG^2 + BF^2 + EF^2 \space \cdots (6)$ .

By the figure, we know that $AG= BF, \space FC = DG$ , then we can rewrite the equation $(3)$ into $AE^2 + EC^2 = BF^2 + EG^2 + EF^2 + DG^2 \space \cdots (7)$ .

We can clearly see that $(7) = (6)$ , hence it is $\color{#3D99F6} \boxed{ \color{#D61F06} \text{True}}$ .