Given a square $ABCD$ and a line $g$. The vertices $A', B', C', D'$ is a projection of every vertex $A, B, C, D$ on $g$. Prove that $(A'C') ^2 + (B'D')^2 = 2(AB)^2$
I tried to draw a vertical line, so the projection is not hard to imagine. But, it keeps me confuse what is the projection here means. By the way, could you help me please?
Let square inscribed in unit circle without losing generality. $A=(\cosθ,\sinθ),B=(\cos(θ+90°),\sin(θ+90°)),C=(\cos(θ+180°),\sin(θ+180°)),D=(\cos(θ+270°),\sin(θ+270°))$.
$$A'C'^2+B'D'^2=(\cosθ-\cos(θ+180°))^2+(\cos(θ+90°)-\cos(θ+270))^2$$ $=4\cos^2θ+4\sin^2θ=4=2AB^2$