In the circle below let the two chords be called $C_1$ and $C_2$, and their intersection be some point that is not the center. The chord power theorem tell us that $a \cdot b = c \cdot d$. I am interested in knowing if there is some similar result for the area of the divided up parts of the circle as well. That is, is there some sort of meaningful relationship between the colored areas? For example, what is the ratio of the total red area to the total area of the circle? What is the ratio of the large red area to the small red area? What information is needed to calculate these/how could I calculate them if I had all the required information?

I have not done any geometry for many years, and I am unsure on how to approach these questions. I was trying to find the ratio between the two red areas, and my idea was to divide the red areas up so that each one consists of a segment and a triangle. Then the two triangles are similar to each other, but I was not sure how to find the area of the segmented areas that remain.Circle


The area of the upper red portion is

$$A_1=\frac{r^2}{2}\left(2\sin^{-1}\left(\frac{\sqrt{b^2+c^2-2bc\cos\phi}}{2r}\right) -\sin\left(2\sin^{-1}\left(\frac{\sqrt{b^2+c^2-2bc\cos\phi}}{2r}\right)\right)\right)+\frac{bc\sin\phi}{2}$$

This is due to the sum of the area of the triangle and segment bound by $b,c$ and the arc. The last term is the triangle and the first (larger) term is the segment based only on $r,b,c,\phi$ where $\phi$ is the angle between segments $b$ and $c$. Likewise, the are of the lower red portion is

$$A_2=\frac{r^2}{2}\left(2\sin^{-1}\left(\frac{\sqrt{a^2+d^2-2ad\cos\phi}}{2r}\right) -\sin\left(2\sin^{-1}\left(\frac{\sqrt{a^2+d^2-2ad\cos\phi}}{2r}\right)\right)\right)+\frac{ad\sin\phi}{2}$$

Same $\phi$ for both because of they are vertical angles. I'm sure that in conjunction with the identity $ab=cd$ that you started with, some kind of relationship between the two can be made.


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