Quadrature of Parabola

Theorem
Let $$T$$ be a parabola.

Consider the parabolic segment bounded by an arbitrary chord $$AB$$.

Let $$C$$ be the point on $$T$$ where the tangent to $$T$$ is parallel to $$AB$$.

Let

Then the area $$S$$ of the parabolic segment $$ABC$$ of $$T$$ is given by:
 * $$S = \frac 4 3 \triangle ABC$$

Proof
We consider WLOG the parabola $$y = a x^2$$.

Let $$A, B, C$$ be the points:
 * $$A = \left({x_0, a x_0^2}\right)$$
 * $$B = \left({x_2, a x_2^2}\right)$$
 * $$C = \left({x_1, a x_1^2}\right)$$


 * [[File:ParabolaQuadrature2.png]]

The slope of the tangent at $$C$$ is given by using:
 * $$\frac{\mathrm{d}{y}}{\mathrm{d}{x}} 2 a x_1$$

which is parallel to $$AB$$.

Thus:
 * $$2 a x_1 = \frac {ax_0^2 - a x_2^2} {x_0 - x_2}$$

which leads to
 * $$x_1 = \frac {x_0 + x_2} 2$$

So the vertical line through $$C$$ is a bisector of $$AB$$, at point $$P$$.

Now, complete the parallelogram $$CPBQ$$.

Also, find $$E$$ which is the point where the tangent to $$T$$ is parallel to $$BC$$.

By the same reasoning, the vertical line through $$E$$ is a bisector of $$BC$$, and so it also bisects $$BP$$ at $$H$$.

Next:

$$ $$ $$ $$

At the same time:

$$ $$ $$

So $$QB = 4 FE = FH$$ and because $$CB$$ is the diagonal of a parallelogram, $$2 FE = 2 EG = FG$$.

This implies that $$2 \triangle BEG = \triangle BGH$$ and $$2 \triangle CEG = \triangle BGH$$

So $$\triangle BCE = \triangle BGH$$ and so as $$\triangle BCP = 4 \triangle BGH$$ we have that:


 * $$BCE = \frac {\triangle BCP} 4$$

A similar relation holds for $$\triangle APC$$:


 * [[File:ParabolaQuadrature1.png]]

... so it can be seen that $$\triangle ABC = 4 \left({\triangle ADC + \triangle CEB}\right)$$.

Similarly, we can create four more triangles underneath $$\triangle ADC$$ and $$\triangle CEB$$ which are $$\frac 1 4$$ the area of those combined, or $$\frac 1 {4^2} \triangle ABC$$.

This process can continue indefinitely.

So the area $$S$$ is given as:
 * $$S = \triangle ABC \left({1 + \frac 1 4 + \frac 1 {4^2} + \cdots}\right)$$

But from Sum of Geometric Progression it follows that:


 * $$S = \triangle ABC \left({\frac 1 {1 - \frac 1 4}}\right) = \frac 4 3 \triangle ABC$$

Historical Note
This proof was given by Archimedes in his book Quadrature of the Parabola, except that he used a different technique to prove that $$\triangle ADC + \triangle CEB = \frac {\triangle ABC} 4$$.