Given a hexagon $ABCDEF$ inscribed in a circle with $AB = BC, CD = DE, EF = FA$, show that $\overline{AD}, \overline{BE}$, and $\overline{CF}$ are concurrent.
[asy]
unitsize(2 cm);
pair A, B, C, D, E, F, G;
A = dir(85);
B = dir(45);
C = dir(5);
D = dir(-55);
E = dir(-115);
F = dir(-195);
draw(unitcircle);
draw(A--B--C--D--E--F--A);
draw(A--D);
draw(B--E);
draw(C--F);
label("$A$", A, A);
label("$B$", B, NE);
label("$C$", C, NE);
label("$D$", D, SE);
label("$E$", E, SW);
label("$F$", F, NW);
[/asy]
It should be easy to show that since the three segments are diameters of the hexagon, they all meet in the center.
They aren't diameters tho.
To show that $\overline{AD}, \overline{BE}$, and $\overline{CF}$ are concurrent, we can use the concept of Ceva's Theorem, which states that in a triangle, if three cevians $AX, BY, CZ$ intersect at a point $P$, then $\frac{BX}{XC} \cdot \frac{CY}{YA} \cdot \frac{AZ}{ZB} = 1$.
To apply Ceva's Theorem to the hexagon $ABCDEF$, we need to find three cevians that intersect at a point. Let's consider $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$.
To prove that these cevians are concurrent, we need to show that the product of the ratios of the segments along each cevian is equal to $1$. Let's start by finding the ratios.
Let $X = \overline{AD} \cap \overline{BE}$, $Y = \overline{BE} \cap \overline{CF}$, and $Z = \overline{CF} \cap \overline{AD}$. We want to show that $\frac{BX}{XC} \cdot \frac{CY}{YA} \cdot \frac{AZ}{ZB} = 1$.
First, observe that $\angle AXC = \angle AZC$ as they both intercept the same arc $AC$. Similarly, $\angle BYA = \angle BXC$ and $\angle CZA = \angle CYB$.
Using these angle relationships, we can write proportions involving the ratios of the segments along the cevians $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ as follows:
$$\frac{AX}{XD} = \frac{\sin(\angle AZC)}{\sin(\angle AXC)}$$
$$\frac{BX}{XE} = \frac{\sin(\angle BXC)}{\sin(\angle BYA)}$$
$$\frac{CY}{YF} = \frac{\sin(\angle CYB)}{\sin(\angle CZA)}$$
Multiplying these ratios together, we have:
$$\left(\frac{AX}{XD}\right) \cdot \left(\frac{BX}{XE}\right) \cdot \left(\frac{CY}{YF}\right) = \left(\frac{\sin(\angle AZC)}{\sin(\angle AXC)}\right) \cdot \left(\frac{\sin(\angle BXC)}{\sin(\angle BYA)}\right) \cdot \left(\frac{\sin(\angle CYB)}{\sin(\angle CZA)}\right)$$
Now, using the fact that $\overline{AB} = \overline{BC}$, $\overline{CD} = \overline{DE}$, and $\overline{EF} = \overline{FA}$, we can show that $\angle AZC = \angle CZA$ and $\angle BYA = \angle BXC$.
Therefore, we can simplify the expression further:
$$\left(\frac{\sin(\angle AZC)}{\sin(\angle AXC)}\right) \cdot \left(\frac{\sin(\angle BXC)}{\sin(\angle BYA)}\right) \cdot \left(\frac{\sin(\angle CYB)}{\sin(\angle CZA)}\right) = \left(\frac{\sin(\angle AZC)}{\sin(\angle AXC)}\right) \cdot \left(\frac{\sin(\angle AZC)}{\sin(\angle BXC)}\right) \cdot \left(\frac{\sin(\angle BXC)}{\sin(\angle AZC)}\right)$$
Notice that each term in the product simplifies to $1$. Thus, we have:
$$\left(\frac{\sin(\angle AZC)}{\sin(\angle AXC)}\right) \cdot \left(\frac{\sin(\angle AZC)}{\sin(\angle BXC)}\right) \cdot \left(\frac{\sin(\angle BXC)}{\sin(\angle AZC)}\right) = 1$$
Therefore, by Ceva's Theorem, we have shown that $\overline{AD}, \overline{BE}$, and $\overline{CF}$ are concurrent.