Irish Mathematical Olympiad

Let $\angle BAC = 2\alpha$, $\angle CBA = 2\beta$ and $\angle ACB = 2\gamma$. Assume first that $2\alpha = \angle BAC = 60^\circ$. This implies $2\beta + 2\gamma = 120^\circ$, i.e. $\beta + \gamma = 60^\circ$. Hence, $\angle DIE = \angle BIC = 120^\circ$. Therefore, $\angle BAC + \angle DIE = 180^\circ$ and the quadrilateral EIDA is cyclic. As $AI$ bisects $\angle BAC$, the chords $EI$ and $DI$ subtend angles of $30^\circ$ at the circumference of the circumcircle of EIDA. This implies $|IE| = |ID|$.

Conversely, assume $|IE| = |ID|$. The bisector $BD$ divides $CA$ in the ratio $|AB| : |BC|$. This can easily be seen from the sine rule for the two triangles $\triangle BDA$ and $\triangle BCD$ and using that $\sin(180^\circ - x) = \sin(x)$.
Let $|BC| = a$, $|CA| = b$ and $|AB| = c$. From $\frac{|CD|}{|DA|} = \frac{a}{c}$ and $|CD| + |DA| = b$ we obtain $|DA| = \frac{bc}{a+b}$. Similarly we get $|AE| = \frac{bc}{a+b}$. Because $|CA| > |AB|$ by assumption, we have $b > c$ and so $\frac{bc}{a+c} > \frac{bc}{a+b}$, hence $|DA| > |AE|$.
Let $D'$ be the reflection of $D$ in $AI$. Since $|DA| > |AE|$, $D'$ will lie between $E$ and $B$ on $AB$. Then $\triangle AID \equiv \triangle AID'$, hence $|ID'| = |ID|$ and $\angle ID'A = \angle ADI = 2\gamma + \beta$. Since $|IE| = |ID|$ we have $|IE| = |ID'|$ from which we get $\angle D'EI = \angle ID'A = 2\gamma + \beta$. From $\angle IEA = 2\beta + \gamma$ we obtain now
$$180^\circ = \angle IEA + \angle D'EI = 2\beta + \gamma + 2\gamma + \beta = 3(\beta + \gamma),$$
which implies $\beta + \gamma = 60^\circ$. Since $\alpha + \beta + \gamma = 90^\circ$, we get $\alpha = 30^\circ$ and so $\angle BAC = 2\alpha = 60^\circ$.