75th Romanian Mathematical Olympiad

Let $\{D\}=BI\cap AC$. Then $\angle BAD=\angle ABD=40^{\circ }$, so $△ABD$ 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 $△IAB\equiv △MBA$ (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 $△AMI$, $D\in (IM)$. 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 $△ADM\equiv △BDC$ (A.S.A.), whence $AM=BC$. Now $△AMI$ 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 $△ABM$ 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 $△AMI\equiv △BAC$ (A.S.A.), which implies $AI=BC$.

Fourth construction. Take $M$ so that $B\in (MC)$ and $△ACM$ is equilateral. Then relations $CA=CM$, $\angle ICA=\angle ICM=30^{\circ }$ and $IC$ common side lead to $△CIA\equiv △CIM$ (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 $△MAB\equiv △CAE$ (A.S.A.), hence $MB=EC$, which implies $ME=BC$ (6). From $△AEM$, $\angle AEM=180^{\circ }-60^{\circ }-40^{\circ }=80^{\circ }$, so, in $△MIE$, $\angle MIE=180^{\circ }-20^{\circ }-80^{\circ }=80^{\circ }=\angle MEI$, hence $△MIE$ is isosceles, whence $MI=ME$ (7). Relations (5), (6) and (7) yield $IA=IM=ME=BC$.