Somemathhere

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