# Proving a result about a triangle, one of its altitude and its circumcircle using complex numbers

Consider the following figure:

Let $A(z_1),B(z_2),C(z_3),E\equiv P(z),O(\mathtt{0})$Q(-z), I need to prove BQ=AC, I can prove it anyways but using complex numbers. Anyways what I tried is as follows: $BQ=|z_2+z|,AC+|z_1-z_3|$ $$\arg\left(\frac{z+z_1}{z_2-z_3}\right)=\arg\left(\frac{z-(-z_1)}{z_2-z_3}\right)=0$$ $$\frac{z_2-(-z)}{z_1-z_3}=e^{i2C}\frac{BQ}{AC}$$

• You have labeled the antipode of $Q$ in the diagram as $E,$ whereas if $Q(-z)$ your saying $P(z)$ would mean the antipode of $Q$ is $P.$ – coffeemath Dec 5 '14 at 17:48
• @coffeemath is it done now? – RE60K Dec 5 '14 at 17:49

Let $a$, $b$ and $c$ be the complex numbers corresponding to the points $A,B,C$. W.l.o.g. let them lie on the unit circle. Then, one easily gets $e=\frac{bc}{a}$ and thus $q=-\frac{bc}{a}$. Now, if you want to prove $BQ=AC$, we shall consider the term $t=\frac{q-b}{a-c}$ and prove that $|t|=1$, i. e. $t \overline{t}=1$. But indeed this can be easily verified by plugging in the equations for $t$ and $q$ and using the unit circle, i.e. $a\overline{a}=b\overline{b}=c\overline{c}=1$.