<s>[INST] Prove $OD^2+OG^2=3AB^2$ 
$AOB$ is cicular sector of $90^{\circ}$.
$C$ is a point on $\stackrel \frown {AB}$.
$ACDE$ and $CBFG$ are squares.
Prove $OD^2+OG^2=3AB^2$
My attempt :
$OA=OC=OB=r$
$O$ is central angle so :
$$\angle O_{1}+\angle O_{2}=\stackrel \frown {BC}=m \hspace{20pt}\angle O_{3}+\angle O_{4}=\stackrel \frown {AC}=n$$
$\angle A_{2}$ and $\angle B_{2}$ are inscribed angles so :
$$\angle A_{2}=\frac{\stackrel \frown {BC}}{2}=\frac{m}{2} \hspace{20pt}\angle B_{2}=\frac{\stackrel \frown {AC}}{2}=\frac{n}{2}$$
in $\triangle OCB$ :
$$\angle OCB=180-(\angle O_{1}+\angle O_{2})-(\angle B_{1}+\angle B_{2})=180-(m)-(45+\frac{n}{2})=135-(m+\frac{n}{2})$$
in $\triangle OCA$ :
$$\angle OCA=180-(\angle O_{3}+\angle O_{4})-(\angle A_{1}+\angle A_{2})=180-(n)-(45+\frac{m}{2})=135-(n+\frac{m}{2})$$
so :
$$\angle OCG=\angle OCB+90=225-(m+\frac{n}{2})=x$$
$$\angle OCD=\angle OCA+90=225-(n+\frac{m}{2})=y$$
after that I applied Law of cosines :
$$OG^2=r^2+CG^2-2r*CG*\cos{x}$$
$$OD^2=r^2+CD^2-2r*CD*\cos{y}$$
And I stuck here.
 [/INST] Hint: Note that $CG = CB$ and $CD = AC$ so:
$CG^2 + CD^2 = CB^2 + AC^2 = AB^2 +2CB·AC·\cos 135 = r^2 + 2CB·AC·\cos 135$.
All you have to do now is to prove that:
$r·CB\cos x + r·AC\cos y = AC·CB·\cos 135$.
Note also that $x = 90 + \angle OCA$ so, applying the formula of cosine of sum of angles: $\cos x = -\sin \angle OCA$. Equivalently, $\cos y = -\sin \angle OCB$.
</s>