Irska

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 $△BDA$ and $△BCD$ 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 $D^{\prime }$ be the reflection of $D$ in $AI$. Since $|DA|>|AE|$, $D^{\prime }$ will lie between $E$ and $B$ on $AB$. Then $△AID\equiv △AI{D}^{\prime }$, hence $|I{D}^{\prime }|=|ID|$ and $\angle I{D}^{\prime }A=\angle ADI=2\gamma +\beta$. Since $|IE|=|ID|$ we have $|IE|=|I{D}^{\prime }|$ from which we get $\angle D^{\prime }EI=\angle I{D}^{\prime }A=2\gamma +\beta$. From $\angle IEA=2\beta +\gamma$ we obtain now $$180^{\circ }=\angle IEA+\angle D^{\prime }EI=2\beta +\gamma +2\gamma +\beta =3(\beta +\gamma ),$$ which implies $\beta +\gamma =60^{\circ }$. Since $\alpha +\beta +\gamma =90^{\circ }$, we get $\alpha =30^{\circ }$ and so $\angle BAC=2\alpha =60^{\circ }$.