Romanian Mathematical Olympiad

Let $\{D\} = BI \cap AC$. Then $\angle BAD = \angle ABD = 40^\circ$, so $\triangle ABD$ is isosceles, yielding $AD = BD$ (1).

First construction. Draw the bisector $BM$ of $\angle DBA$, with $M \in AC$. Then $\angle ABM = \angle IAB = 20^\circ$ and $AB$ is a common side, hence $\triangle IAB \equiv \triangle MBA$ (A.S.A.), whence $AI = BM$.

Then relations $\angle CBM = \angle CBD + \angle DBM = 60^\circ$ and $\angle MCB = 60^\circ$ show that triangle $MBC$ is equilateral, therefore $BC = BM = AI$.

Second construction. Take the equilateral triangle $\triangle AMI$, $D \in (IM)$. Since $\angle MAD = \angle MAI - \angle IAD = 60^\circ - 20^\circ = 40^\circ$, $\angle MAD = \angle DBC$ (2). From (1), (2) and $\angle AMD = \angle DCB = 60^\circ$ follows $\triangle ADM \equiv \triangle BDC$ (A.S.A.), whence $AM = BC$. Now $\triangle AMI$ equilateral implies $AI = AM$, hence $AI = BC$.

Third construction. Let $AM \parallel BC$, $M \in BI$. Then $\angle AMB = \angle MBC$ (alternate angles), implying $\angle ABM = \angle AMB = 40^\circ$, hence $\triangle ABM$ is isosceles, whence $AB = AM$ (3). From $\angle CAM = \angle ACB = 60^\circ$ (alternate angles) follows $\angle IAM = 20^\circ + 60^\circ = 80^\circ = \angle ABC$ (4). Relations (3), (4) and $\angle AMI = \angle BAC = 40^\circ$ give $\triangle AMI \equiv \triangle BAC$ (A.S.A.), which implies $AI = BC$.

Fourth construction. Take $M$ so that $B \in (MC)$ and $\triangle ACM$ is equilateral. Then relations $CA = CM$, $\angle ICA = \angle ICM = 30^\circ$ and $IC$ common side lead to $\triangle CIA \equiv \triangle CIM$ (S.A.S.), whence $IA = IM$ (5) and $\angle IAC = \angle IMC = 20^\circ$.

Let $\{E\} = AI \cap BC$. From $\angle MAB = \angle MAC - \angle BAC = 20^\circ$, $AM = MC$ and $\angle AMB = \angle ACE = 60^\circ$ follows $\triangle MAB \equiv \triangle CAE$ (A.S.A.), hence $MB = EC$, which implies $ME = BC$ (6). From $\triangle AEM$, $\angle AEM = 180^\circ - 60^\circ - 40^\circ = 80^\circ$, so, in $\triangle MIE$, $\angle MIE = 180^\circ - 20^\circ - 80^\circ = 80^\circ = \angle MEI$, hence $\triangle MIE$ is isosceles, whence $MI = ME$ (7). Relations (5), (6) and (7) yield $IA = IM = ME = BC$.