The following is my own attempt at proving the Basic Proportionality Theorem. Corrections and suggestions are welcome.
To prove: $\frac{\mathrm{\overline{A'B'} }}{\mathrm{\overline{AB} }}= \frac{\mathrm{\overline{B'C'} } }{\mathrm{\overline{BC} } }=\frac{\mathrm{\overline{A'C'}} }{\mathrm{\overline{AC}} }= \mathrm{K} $, where $\mathrm{K} $ is a constant.
Proof: In the above figure, $\Delta{\mathrm{A'B'C'} }$ is inscribed in the larger circle of radius $\mathrm{R'} $ and $\Delta{\mathrm{ABC} }$ is inscribed in the smaller circle of radius $\mathrm{R} $. Here, $\Delta{\mathrm{A'B'C'} }$ is the enlarged version of $\Delta{\mathrm{ABC} }$.
A well-known result from elementary geometry relates the arc length $l$ of a circle, the angle $\mathrm{\theta} $ subtended by the arc and the radii $\mathrm{R} $ joining the ends of the arc using the equation, $l=\mathrm{R\cdot\theta} $.
From the figure and the above result, we see that $\mathrm{arc A'B'} = \mathrm{R'\theta} $ and $\mathrm{arc AB} = \mathrm{R} \theta$, where $\mathrm{\theta} $ is $\angle{\mathrm{A'OB'}}$. Taking the ratios we get, $$\frac{\mathrm{arcA'B'}} {\mathrm{arcAB}} = \frac{\mathrm{R'}} {\mathrm{R}} $$
Similarly, $$\frac{\mathrm{arcB'C'}} {\mathrm{arcBC}} = \frac{\mathrm{R'}} {\mathrm{R}}$$ and $$\frac{\mathrm{arcA'C'}} {\mathrm{arcAC}} = \frac{\mathrm{R'}} {\mathrm{R}} $$
The diameter whose length is $\mathrm{D}$ is the longest chord of a circle. The other chords are scaled-down versions of the diameter. Mathematically, $$\mathrm{P} = \mathrm{B} \cdot \mathrm{D} $$
Where $\mathrm{B} $ is a scaling factor—whose value is between $0$ and $1$—and where $\mathrm{P} $ is the chord length.
Let $\mathrm{P} $ in figure 2 be the diameter length. The length of the diameter changes from $\mathrm{D} $ to $\mathrm{D'} $. It is evident that the scaling factor is the same($\mathrm{B} =1$) in the larger circle and the smaller circle, for the ratio of the diameter lengths in each circle is $1$.
Now let $\mathrm{P} $ be the length of a chord that subtends an angle of $\frac{\pi} {3}$ radians at the center of a circle(see figure 3). The chord length is equal to the radius length of the circle, for the chord is a part of an equilateral triangle. As the chord length goes from $\mathrm{P} $ to $\mathrm{P'} $, it is, once again, evident that the scaling factor is the same($\mathrm{B} =\frac{1}{2}$) in the larger circle and the smaller circle, for the chord length in each circle is half of the diameter length.
We see a pattern. The pattern tells us that for a fixed $\theta$, the scaling factor is the same in every concentric circle. In other words, the scaling factor is independent of the radius length and is a constant for a constant $\theta$.
Following the pattern we have, $$\mathrm{P} \propto \mathrm{D} $$
This implies, $$\mathrm{P} \propto \mathrm{R} $$
This result and the result $l=\mathrm{R\cdot\theta} $(for constant $\mathrm{\theta} $) together imply, $$\mathrm{P} \propto l$$
Returning to F figure 1 we see, $$\overline{\mathrm{A'B'} }= \mathrm{C} \cdot \mathrm{arcA'B'}$$
In the same sector, $$\overline{\mathrm{AB}} = \mathrm{C} \cdot \mathrm{arcAB}$$
Where $\mathrm{C}$ is a constant. Taking the ratios we get, $$\frac{\mathrm{arcA'B'}} {\mathrm{arc AB}} = \frac{\overline{\mathrm{A'B'}} }{\overline{\mathrm{AB}} } $$
Similarly, $$\frac{\mathrm{arcB'C'}} {\mathrm{arc BC}} = \frac{\overline{\mathrm{B'C'} } }{\overline{\mathrm{BC}} } $$
and
$$\frac{\mathrm{arcA'C'}} {\mathrm{arc AC}} = \frac{\overline{\mathrm{A'C'}} }{\overline{\mathrm{AC}} } $$
The above results convey, $$\frac{\mathrm{\overline{A'B'} }}{\mathrm{\overline{AB} }}= \frac{\mathrm{\overline{B'C'} } }{\mathrm{\overline{BC} } }=\frac{\mathrm{\overline{A'C'}} }{\mathrm{\overline{AC}} }= \mathrm{\frac{R'}{R}} $$
Where $\frac{\mathrm{R'}} {\mathrm{R}} = \mathrm{K}$, a constant. This concludes the proof.