IMO shortlist, 1998

G7. $ABC$ is a triangle with $\angle ACB = 2 \angle ABC$. $D$ is a point on the side $BC$ such that $DC = 2 BD$. $E$ is a point on the line $AD$ such that $D$ is the midpoint of $AE$. Show that $\angle ECB + 180 = 2 ∠EBC$.

 

Solution.  Put $D$ in the center. We can rewrite everything in terms of $a$ - the number corresponding to $A$ and $b$ - the number corresponding to $B$. Thus $C$ is $-2b$, E is $-a$.
The relation between the angles at $B$ and $C$ is easily reduced to the following equation:
$$\frac{(a-b)^2}{|a-b|^2}\frac{a+2b}{|a+2b|}=\frac{b^3}{|b|^3}.$$
Squaring and representing modules with conjugates, we obtain
$$\frac{(a-b)^2}{(\bar a-\bar b)^2}\frac{a+2b}{\bar a+2\bar b}=\frac{b^3}{\bar b^3}.$$
Clearing the denominator and setting everything to the left we obtain
$$a^3\bar b^3-3a\bar b^3 b^3-\bar a^3b^3+3\bar a\bar b^2 b^3=0\ (*)$$
Similarly, the equality we want to proof is equivalent to
$$\left(\frac{\frac{-b}{|-b|}}{\frac{-a-b}{|-a-b|}}\right)^2=(-1)
\left(\frac{\frac{-a+2b}{|-a+2b|}}{\frac{b}{|b|}}\right)$$
Both of the arguments of the complex numbers on both sides are between 180 and 360, so squaring is equivalent transformation here. As above we obtain
$$\frac{(a+b)^2}{(\bar a+\bar b)^2}\frac{a-2b}{\bar a-2\bar b}=\frac{b^3}{\bar b^3}.$$
Everything on the left and expand - we obtain
$$-a^3\bar b^3+3a\bar b^3 b^3+\bar a^3b^3-3\bar a\bar b^2 b^3=0.$$
But this is just the same as $(*)$.

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