The problem:
Suppose $AB, CD, EF, GH$ are four equally inclined (angle of $45^\circ$) chords of a circle concurrent at an interior point $P$ of the circle, dividing the circle into 8 parts. Show that the sum of the areas of sector-like regions $HPA, CPE, GPB, DPF$ is equal to the sum of the areas of the remaining regions $APC, EPG, BPD, FPH$.
Figure:
My solution:
We use the area of curve from point $P$, and denote the distance $PX$ by $r(\theta)$ if $\angle XPA = \theta$ and $X$ lies on the circle. Then,
$$\begin{align}2(\mathrm{Ar}(APC)+\mathrm{Ar}(EPG)+\mathrm{Ar}(BPD)+\mathrm{Ar}(FPH)) &= \int_0^{\pi/4} r(\theta)^2 \mathrm d\theta + \int_{\pi/2}^{3\pi/4} r(\theta)^2 \mathrm d\theta + \int_{\pi}^{5\pi/4} r(\theta)^2 \mathrm d\theta + \int_{3\pi/2}^{7\pi/4} r(\theta)^2 \mathrm d\theta \\ &= \int_0^{\pi/4} r(\theta)^2 + r(\theta + \pi/2)^2+ r(\theta + \pi)^2 + r(\theta+3\pi/2)^2 \mathrm d\theta \\ &= \int_0^{\pi/4} 4R^2 \mathrm d\theta \\ &= \pi R^2 \end{align}$$ where $R$ is the radius of the circle, and the second-last equality uses Archimedes' 11th proposition.
Question:
This problem comes from an olympiad book, and the book has nowhere mentioned integrals (not even derivatives), or used calculus, so I feel there should be an alternate and more elementary approach. How should I go about solving this without calculus? Also, I don't think the book intended a dissection approach. Probably standard Euclidean or co-ordinate geometry only.