Final Round

a. In 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 $$\angle BAC=\angle BCE.\square$$

b. Triangles $△ACE$ and $△CBE$ 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 $$|AD|=|AB|-|BD|=5-2=3\text{and}|ED|=6.$$ We obtain that $2|AD|=2\cdot 3=6=|ED|$. $\square$