[Intro Music]
What is the area of the section of a parabola,
formed by the chord joining any two points on the curve?
Archimedes, the Greek mathematician, gave an elegant answer to this question,
 without the knowledge of calculus.
His answer was that, the area of the segment is equal to
two thirds of the area of the triangle formed by the two tangents 
intersecting at a point.
Let the two points be A and B 
This is the area we are supposed to find.
And the point of intersection of the two tangents is P.
Draw a line parallel to the axis of symmetry of the parabola.
Note that the axis of symmetry of the parabola passes through the vertex of it.
In the case considered here, the line PM and the axis are aligned.
An interesting property of a parabola is that the line passing through P, 
parallel to the axis of symmetry bisects the chord AB.
Do try to prove this yourselves, but the proof is in the description. 
The tangent at Q intersects the tangents at A and B at A1 and B1 respectively.
Now, when we look at the arc between points Q and A, it is also a part of the parabola, 
and the tangents at Q and A intersect at A1.
So as we saw before, a line parallel to the axis the parabola
passing through A1 bisects the chord AQ.
Now we know that lines PQ and A1L are parallel, 
and point L is the midpoint of AQ, 
so this means that A1's got to be the midpoint of the line AP.
With a similar line of argument, it can be shown that BB1 is equal to B1P.
So A1B1 is actually parallel to the chord AB and is half way in-between.
Now, it can be shown that the triangles PBM and PB1Q 
are similar as two sides are proportional and they have common angle. 
Similarity implies that the line PQ is half of PM, 
which also implies that PQ is equal to QM.
So the height and base of the triangle PB1A1 is half of that of PBA.
which means the area of PBA is four times the area of PB1A1.
And similarly the height of triangle QBA is half of that of the triangle PBA.
Therefore the area of triangle PBA is twice the area of QBA.
And from these two equations, it can be concluded that the area of triangle QBA
is twice the area of PB1A1.
So now, we can do all this, to this arc on the left side and arrive at the same result. 
Similarly, to the arc on the right.
Now we know that all the triangles inside the parabola 
are twice of that of the ones outside.
So... We've gotten somewhere nearer to the answer. 
At the limiting case, meaning applying the same arguement to all those little arcs, 
We can prove that the parabolic segment is equal to
two times the remaining area.
We know that the Archimedes triangle PBA 
is a sum of this parabolic segment and the remaining part
As we just proved, this remaining area is half of the parabolic segment. 
Adding and rearranging the equation we arrive at the answer, 
that is the area of the parabolic segment is equal to 
two thirds of the Archimedes triangle PBA.
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