Ancient Greeks were very advanced in plane geometry, and apart from Euclid and Pythagoras,
Heron is probably the most heard of, known for his formula of calculating area of triangles:
sqrt (s(s-a)(s-b)(s-c)), where s is half the perimeter of triangle, and a,b,c denotes the
three side lengths. You might be surprised that I said plane geometry when introducing
Heron’s formula, because it is usually put in the trigonometry section of a textbook.
This is only because it is most easily justified using trigonometry, but Heron originally used
plane geometry. Today we are going to look at 4 different proofs in detail, and see which
of those is your favorite.
The first proof is the most seen, because it is probably also used in textbooks. Area
of a triangle is half base times height. We can let this side with length a to be the
base, and the height can then be represented as b times sin C, so the area is half ab sinC,
which is the first result we need for this proof. Next, we have cosine law. If you want
a proof of cosine law, stick to the end of the video, because I don’t want to interrupt
the flow and side track too much. Then we have one final identity: the sum of sine squared
and cosine squared of any angle is 1. So armed with these three facts, we are ready to turn
on the algebra machine. So this could be done with cosine law. Another proof only relies
on Pythagoras’ theorem. Consider this triangle, where we split the segment of length a into
two parts by the height. Based on Pythagoras’ theorem, we have these two relations. Subtract
the second equation from the first, we have this equation, which can be rearranged as
follows. Now the area is actually half a times h. Now we get this formula, and the algebra
from now on would be similar to what we have done above, so I don’t bore you with it.
These two proofs rely very heavily on algebra, so are there any other ways to prove Heron’s
formula without these heavily machinery? These two methods presented below would be more
heavily reliant on geometry instead.
These two proofs both in turn rely on these results. First, we have a novel way to calculate
the area of the triangle using this diagram, where I here is the incentre of the large
triangle. To save space, we use the notation that S with the underscript XYZ meaning the
area of the triangle stated. Then we have the large triangle being the three smaller
triangles combined, and we can apply half base times height to each of the smaller triangles,
and get that the area can also be calculated by the product of r, the inradius, and s,
the semiperimeter. Another result we need is the geometric meaning of s-a, s-b, and
s-c. In this diagram, line segments with the same colour are of the same length. If you
really want, here is a simple justification. Pause the video if you want. From the diagram,
it is now easy to see that on the perimeter, red, green and blue segments have appeared
twice. So they add up to the perimeter, and then divide it by 2. Now, if we concentrate
on side AB, we have red and green segments adding up to c, so blue segment has the length
s-c. Similarly, the red segment has length s-a, while the green one has length s-b. A
good mnemonic is that the segments from A would have length s-a, from B would have length
s-b, and from C would have length s-c.
Before we start with Heron’s original proof, here is just two more results specific to
this third proof, where the proof of these results will be at the last part of the video.
If any quadrilateral fulfils either of the criteria, it can be inscribed in a circle,
that is, cyclic. Conversely, if any quadrilateral is cyclic, it must have these two properties.
We name these two results A and B respectively for future reference. Now we are ready for
the proof. First off, we have this colorful diagram, where the angles at the centre are
of the same size if they are colored the same, which can also be easily justified using congruent
triangles shown previously. So now, considering the angles around the point I, we have two
angles of each color, and so red, greeen, and blue angles add up to 180 degrees. We
now associate the red angle to be angle AIE, and combined green and blue angle to be angle
BIC. Then we can conclude that these two angles add up to 180 degrees.
Ok, now we deliberately create a cyclic quadrilateral, by constructing these perpendicular lines,
which intersect at H, and then denote the intersection of IH and BC to be G. Then the
quadrilateral ICHB would be cyclic, because the two purple angles are the same, and we
have the property A. Now since it is cyclic, these two opposite angles would add up to
180 degrees, due to property B. From what we have discussed, the yellow and red angles
in the middle add up to 180 degrees, so the two red angles are of the same size. Combining
with the two right angles, we can conclude that these two grey triangles are similar.
There are just two more pairs of similar triangles which are easy to identify. Firstly, for these
two triangles, in addition to the right angles, we also have these two vertically opposite
angles, so it is enough to prove the similarity. These two are also similar, because the yellow
and red angles add up to angle BIG, which is by construction 90 degrees, and similarly
in a triangle, the yellow and red angles also add up to 90 degrees, so the two red angles
are of the same size. With the right angle, we can also prove the similarity of these
two triangles.
It might be a lot to take in, so let me summarize what just happened so far: we managed to arrive
at the conclusion that the area is the product of r and s, as well as the side lengths expressed
in terms of s, a, b, and c. Then we prove three pairs of similar triangles. If you don’t
understand how we came up with these, first think over it by yourself. I assure you that
you will gain a much deeper understanding by thinking through this. However, if you
have absolutely no idea then of course you should replay what we have done, and pause
where necessary. We now move on.
From this pair of similar triangles, we have this proportionality of sides. Now since both
ID and IE are the radius of the circle, they are of the same length, so we can replace
ID by IE. Now, we look at another pair of similar triangles. We have another proportion
of sides. We can now replace BC by the symbol a, and from the results we obtained, we have
AE to be s-a. Now we add 1 to both sides, and rewrite 1 in different forms. Now the
sum of GD and CG is, from the diagram, CD, and from the results previously obtained,
it is actually s-b. Now focus on the right side, which is s over s-a. Rearranging, we
have this formula. We are quite close now. From this pair of similar triangles, we have
this proportion of sides. Since ID is the inradius, we replace ID by r, and from the
previous results, we have BD to be s-b. Rearranging, we get this formula. Combining with the result
derived a minute ago, we first multiply both formulae, cancel GD on both sides, then multiply
both sides by s. And now square root both sides. Now, since the area is r times s, we
finally have the Heron’s formula. Now given how complicated this proof is, we really need
to give a round of applause to Heron to come up with the entire proof. But now we can have
much less complicated geometric proof after so many years.
We would still use these two results. First, we construct external angle bisectors of B
and C, which will intersect at a point, which we call I_a. Since angle bisectors are the
collections of points that are equidistant from the two lines that create the angle,
which can be proved pretty easily, that means I_a is equidistant from BF, BC, and CD, and
so also lie on the angle bisector of A, and of course, because of the equal distance,
we can draw a circle just touching the sides. Since we have previously demonstrated that
tangential segments from a point are of the same length, we have the pairs of blue and
purple segments having the same length, and so they add up to length BC. At the same time,
we have AF equal to AD, so we also have this relation. Rearranging, we get the difference
between these lengths. By subtracting the second relation from the first, we have this.
What’s interesting is that we of course have AD as the sum of these two lengths, which
can be found to be s. So we give s its geometric meaning. Now we have this pair of triangles
to be similar, which is pretty evident from the green common angle and the right angles.
So we have this proportion of sides. Since we have the previous result, we turn AE to
s-a, and then we have this formula. For this pair of triangles, we observe that around
vertex C, red and yellow angles appear twice, adding up to 180 degrees, so these two angles
add up to 90 degrees. At the same time, these two angles add up to 90 degrees because of
how angles sum up in a triangle. Combining with the two right angles, we can prove the
similarity of the two triangles. Now, we write down the proportionality of sides, and use
the results we have before, we have this relation. Multiplying these two results we just concluded
using similar triangles, then divide both sides by I_aD, then multiply both sides by
s, and again, observe that the area is actually the square root of the right hand side, so
we have again proved the Heron’s formula.
I promised you guys that I will provide the proof of cosine law, cyclic quadrilateral
properties A and B here, so I made a separate video about it. Check that out if you are
unsure how to derive those. For now, subscribe if you want more of these math videos, and
see you next time.