Prove $OD^2+OG^2=3AB^2$

Question

$AOB$ 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.

Answer 1

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, $\cos y = -\sin \angle OCB$.

Answer 2

By embedding the construction in $\mathbb{C}$ and setting: $$O=0,\quad B=1,\quad A=i,\quad C=e^{i\theta},\quad \theta\in(0,\pi/2)$$ we have: $$G=(B-C)i+C = i+(1-i)e^{i\theta},\qquad D=-i(A-C)+C = 1+(1+i)e^{i\theta},$$ from which $$\bar{G}=-i+(1+i)e^{-i\theta},\qquad \bar{D}=1+(1-i)e^{-i\theta}$$ and: $$ OG^2+OD^2 = G\cdot\bar{G}+D\cdot\bar{D} = 1-i^2+2(1-i^2)=6$$ since all the terms depending on $e^{i\theta}$ or $e^{-i\theta}$ cancel out nicely.