# A Proof without Trigonometric Function

Problem: $A$ and $B$ are squares. Without invoking trigonometric functions, show that the area of triangle $C$ equals that of $D.$

Solution: Introducing a rectangular coordinate system and complex number $V_1, V_2, P, W_1, W_2, Q:$

Let $A_1, A_2$ denote the area of triangle $C$ and $D$ respectively.

We have

$P = V_2-V_1.\quad\quad\quad(1)$

Let $s_1 = \frac{|V_1| + |V_2| + |P|}{2} \overset{(1)}{=} \frac{|V_1| + |V_2| + |V_2 - V_1|}{2}.$

By Heron’s formula derived without invoking trigonometric function (see “An Algebraic Proof of Heron’s Formula“),

$A_1 = \sqrt{s_1(s_1-|V_1|)(s_1-|V_2|)(s_1-|P|)}\overset{(1)}=\sqrt{s_1(s_1-|V_1|)(s_1-|V_2|)(s_1-|V_2-V_1|)}.\quad(*)$

We also have

$W_1 = \bold{i}V_1, \quad W_2 = -\bold{i}V_2\quad\quad\quad(2)$

(see “Treasure Hunt with Complex Numbers“) and,

$Q = W_2-W_1\overset{(2)}{=}-\bold{i}V_2-\bold{i}V_1.\quad\quad\quad(3)$

Similarly,

Let $s_2 = \frac{|W_1| + |W_2| + |Q|}{2} \overset{(2), (3)}{=} \frac{|iV_1| + |-\bold{i}V_2| +|-\bold{i}V_2-\bold{i}V_1|}{2}=\frac{|V_1| + |V_2| +|V_2+V_1|}{2}.$

$A_2 = \sqrt{s_2(s_2-|W_1|)(s_2-|W_2|)(s_2- |Q|)}$

$\overset{(2), (3)}{=}\sqrt{s_2(s_2-|\bold{i}V_1|)(s_2-|-\bold{i}V_2|)(s_2- |-\bold{i}V_2-\bold{i}V_1|)}.$

That is,

$A_2=\sqrt{s_2(s_2-|V_1|)(s_2-|V_2|)(s_2- |V_2+V_1|)}\quad\quad\quad(**)$

It is shown by Omega CAS Explorer that the expression under the square root of $A_1$ is the same as that of $A_2:$

Therefore,

$A_1 = A_2.$

Exercise-1 The two squares with area 25 and 36 ( see figure below) are positioned so that $AB=7.$ Find the area of triangle TSC.