[Problem]
Let C, E, F, and D be points on a circle. Suppose G is the intersection point of CD and EF. Furthermore, $\angle DGE = \angle FGC = 120^{\circ}$, $\angle GHI= \angle GJI= 90^{\circ}$, and $\angle CAE = 60^{\circ}$. Moreover, we know that $|GH| = |GJ| = \sqrt[4]{48}$ If the area of the region CGF is the same with the area of region of EGD that is $30 \, cm^{2}$ find the total area of region of CGE and FGD!
My attempt so far in the problem is that I know $|HI| = |IJ| = \frac{\sqrt[4]{48}}{\sqrt{3}}$ and the triangle AGB is an equilateral triangle. However, I do not know how to use this information to solve the question. Anhe help is much appreciated! Thank you and I am sorry to ask such basic elementary geometry problem!
Best Answer
Since $\angle DGE = \angle FGC = 120^{\circ}$, then we can easily check that $\angle CGE = \angle FGD = 60^{\circ}$.
You have also stated that $\angle CAE = 60^{\circ}$. In this case, point $G$ must be equal to point $A$ (the center of the circle). That's the only possibility that could explain $\angle CGE = \angle CAE = 60^{\circ}$.
So the areas $CGF = CAF = EGD = EAD = 30 cm^2$. With that in mind, we can find the radius of the circle $R$: $$ Area_{CAF} = Area_{EAD} = \cfrac{\theta R^2}{2} = \cfrac{\pi R^2}{6} = 30 cm^2 \therefore R^2 = \cfrac{180}{\pi} \therefore R = \sqrt{\cfrac{180}{\pi}} $$
Finally, the areas $CGE$ (= $CAE$) and $FGD$ (= $FAD$) can be found:
$$ Area_{CGE} = Area_{FGD} = \cfrac{\theta R^2}{2} = \cfrac{\pi R^2}{12} = \cfrac{\pi}{12} \cfrac{180}{\pi} = 15 cm^2 $$