# An Interesting proof in Triangles with Complex Algebra

This is such an interesting problem that we shall discuss here. How can you derive something that is real, using complex geometry and algebra? What really fascinates me about mathematics is that it can relate two completely different things together. Take algebra and plotting for example. For many years, we thought that algebra had little to no geometrical meaning. This was later proven false when Descartes laid the foundations of the Cartesian Plane and the x and y axes. Now you can algebraically interpret shapes such as circles, ellipses, lines, and even parabolas and hyperbolas in the form of algebraic equations in x and y coordinates. When I was checking Mathematics Stack Exchange the other day, I came across a new problem that I found simply fascinating and that is the subject of this blog post here. Today we shall use complex numbers and complex geometry to derive Heron’s Formula. The Original proof by Miles Dillon can be found here, but we will be using a proof that is a bit different. But first, let us talk about what is Heron’s Formula.

# Heron’s Formula: A brief Introduction

Heron’s Formula is an equation that tells you the area of a triangle provided the lengths of the sides of the triangle are known. Using the traditional formula for area = 1/2*base*height can be a bit tricky because in most cases other than right triangles, you need to find the height of the triangle yourself, which can be a tricky task if you do not know trigonometry.

Hence Heron’s Formula is used. If the sides of the triangle are a, b, and c respectively, then Heron’s formula states that:

Where s is the semi-perimeter of the triangle, which is defined as:

Now that we are familiar with Heron’s Formula, let us derive it using complex geometry.

# The Derivation

When we derive the equation, we shall make use of the following diagram.

Let I be the incentre³ of the △ABC. Now the semi-perimeter ‘s’ of △ABC is given by the sum of all sides over two.

Simplifying, we get:

Thus s is:

Now that we have ‘s’ out of the way, we shall focus on angles. Since α, ß and γ go around the circle,

This implies that:

Now let us consider I as the origin, and A, B, and C as separate points (or numbers). If the distance IA is u, the distance IB is v and IC is w, we can say that uvw is:

Taking r out and evaluating the exponents on e we get:

Using the sum of α, ß and γ we have found above, you can arrive at¹:

Using Euler’s Identity,

where k is some constant.

Since -k is a purely real number, the imaginary part of this complex product must be 0.

Multiplying, we get:

This implies that:

Taking the r² out we get:

Now solving for r, we get:

Further Simplification yields:

Since x+y+z=s, we get:

Now since x+y+z=s, x=s-(y+z)=a, z=s-(x+y)=c and y=s-(x+z)=b where a, b, and c are the length of sides opposite to A, B and C respectively.

Thus r becomes:

Now the area of △ABC can also be found by adding up the area of the little triangles inside it². So:

Since △AIE and △AIF are congruent to each other, their area is the same. Similarly, △IEC and △IEF, along with △BFI and △BDI will form equal-area pairs.

So the area becomes:

Using area=1/2*base*times height and taking the two out we get:

Now take the half out:

The half and the 2 cancel out. Taking the r out leaves us with:

Since x+y+z=s, the area now becomes:

Since r is:

Thus our area now becomes:

The s and √s cancel leaving us with:

multiplying √s we get:

Which is the required solution!

Hence we have proved Heron’s Formula using complex numbers!

# Some Important Points

Once Again I have placed some markers in the form of numbers in superscript to highlight points that need to be taken care of. Here I shall explain them.

1: This is a direct consequence of Euler’s Identity.

2: Even though we have simply added up the areas like scalars, however, in some branches of Physics (especially electrostatics, using Gauss’ law and such)Area is considered a vector. Then addition has to be done using the Triangle and Parallelogram Laws of Vector Addition. In such cases, the direction of the area vector is taken to be normally outward to the surface whose area we are taking into consideration, as shown.

3: The Incentre of a triangle is the point where all the angle Bisectors of the triangle meet. If we draw a circle with Radius = the perpendicular distance of incentre from the sides of the triangle, we get a circle which has the three sides as tangents to it.

This is it for this article. Next up, we will try to decipher the LMVT! Have a good day ahead and stay safe!