Consider the following figure:

enter image description here

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: $BQ=|z_2+z|,AC+|z_1-z_3|$ $$\arg\left(\frac{z+z_1}{z_2-z_3}\right)=\arg\left(\frac{z-(-z_1)}{z_2-z_3}\right)=0$$ $$\frac{z_2-(-z)}{z_1-z_3}=e^{i2C}\frac{BQ}{AC}$$

  • $\begingroup$ 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 $P.$ $\endgroup$ – coffeemath Dec 5 '14 at 17:48
  • $\begingroup$ @coffeemath is it done now? $\endgroup$ – 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. $a\overline{a}=b\overline{b}=c\overline{c}=1$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.