Netherlands — Final Round

a.
In triangle $\triangle ADC$, the sum of the angles is $180^\circ$, hence
$$\angle BAC = \angle DAC = 180^\circ - \angle ADC - \angle ACD.$$
Because $CD$ is the angle bisector of $\angle ACB$, we have $\angle ACD = \angle DCB$ and hence the equality above can be rewritten as
$$\angle BAC = 180^\circ - \angle ADC - \angle DCB.$$
Now we use that $\angle ADB$ is a straight angle, hence $\angle EDC = 180^\circ - \angle ADC$. Substituting this yields
$$\angle BAC = \angle EDC - \angle DCB.$$
Because $E$ lies on the perpendicular bisector of $CD$, we have $\angle EDC = \angle ECD$, and the equality becomes
$$\angle BAC = \angle ECD - \angle DCB.$$
Finally, we also see in the picture that $\angle ECD - \angle DCB = \angle BCE$, and hence
$$\angle BAC = \angle BCE. \quad \square$$

b.
Triangles $\triangle ACE$ and $\triangle CBE$ are similar, because $\angle AEC = \angle CEB$ (same angle) and in part (a) we proved that $\angle BAC = \angle BCE$ and hence $\angle CAE = \angle BCE$. This yields
$$\frac{|AE|}{|CE|} = \frac{|CE|}{|BE|}.$$
Using the fact that $|BE| = 4$, we compute
$$|AE| = |AB| + |BE| = 5 + 4 = 9.$$
Substituting this in the ratios above, we obtain
$$\frac{9}{|CE|} = \frac{|CE|}{4},$$
hence $|CE|^2 = 36$ and $|CE| = 6$. Because the perpendicular bisector of $CD$ passes through $E$, we have $|CE| = |DE|$. This yields
$$6 = |CE| = |DE| = |DB| + |BE| = |DB| + 4$$
and hence $|DB| = 2$. Therefore, we conclude that
$$|AD| = |AB| - |BD| = 5 - 2 = 3 \quad \text{and} \quad |ED| = 6.$$
We obtain that $2|AD| = 2 \cdot 3 = 6 = |ED|$. $\square$