I did the following easy exercise in 2020:
Points $A$, $B$, and $C$ are on a circle with centre $O$.
The lines through $O$ parallel to $AC$ and $BC$ intersect the circle at $D$ and $E$.
Then the area of the sector $AOB$ is twice the area of the sector $DOE$.
It is true for conic sections in general:
Points $A$, $B$, and $C$ are on a hyperbola with centre $O$.
The lines through $O$ parallel to $AC$ and $BC$ intersect the hyperbola at $D$ and $E$.
Then the area of the sector $AOB$ is twice the area of the sector $DOE$.
Can we deduce the hyperbola case from the circle case through some geometric transformation?
By applying a suitable affine transformation we can assume that the hyperbola is given by $xy=1$ and $C=(-1,-1)$. Let $F=(1,1)$. It is enough to prove that the area of sector $FOB$ equals twice the area of sector $FOE$. (We prove in a similar way that the area of $AOF$ equals twice the area of sector $DOF$, from which the result follows.)
Let $E=(s,\frac 1s)$. As shown here, the area of sector $FOE$ equals $\ln s$. Similarly, if $B=(t,\frac 1t)$ then the area of sector $FOB$ equals $\ln t$. We have to prove that $\ln t = 2\ln s$, which is equivalent to $t=s^2$.
Note that the line $OE$ has equation $y=\frac{1}{s^2}x$. By assumption, the line $BC$ is parallel to $OE$, hence $BC$ has equation $y=\frac{1}{s^2}x+u$ for some $u$. Since $C=(-1,-1)$, we have $-1=\frac{1}{s^2}\cdot(-1)+u$, hence $u=-1+\frac{1}{s^2}$. We have $\frac 1t = \frac{1}{s^2}t-1+\frac{1}{s^2}$. This is a quadratic equation in $t$ with roots $-1$ and $s^2$. Since $B\neq C$, we have $t\neq -1$, hence $t=s^2$ and we are done.