Saudi Arabia Mathematical Competitions 2012

The bisector of $\widehat{BAC}$ and the perpendicular bisector of side $BC$ intersect at $Q$, the midpoint of arc $\widehat{BC}$ not containing $A$. We have $BQ=IQ$, since $\widehat{QBI}=\widehat{QIB}=90^{\circ }-\frac{1}{2}\hat{C}$. On the other hand, the triangles $BLQ$ and $ALC$ are similar, meaning that $$\frac{BQ}{LQ}=\frac{AC}{LC}.$$ Using the angle bisector theorem in the triangles $ABC$ and $ABL$, we get $$\frac{IQ}{LQ}=\frac{BQ}{LQ}=\frac{AC}{LC}=\frac{AB}{BL}=\frac{AI}{IL}.$$ so $AI\cdot LQ=IQ\cdot IL$, as desired.

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Alternative solution.

Let us use the diagram in the previous solution. We have $$\frac{AI}{IQ}=\frac{K[ABI]}{K[QBI]}=\frac{c\sin \frac{B}{2}}{BQ\sin \frac{A+B}{2}}=\frac{c\sin \frac{B}{2}}{BQ\cos \frac{C}{2}}$$ $$=\frac{2R\sin C\sin \frac{B}{2}}{2R\sin \frac{A}{2}\cos \frac{C}{2}}=\frac{2\sin \frac{B}{2}\sin \frac{C}{2}}{\sin \frac{A}{2}},(1)$$ and $$\begin{array}{rl}\frac{IL}{LQ}& =\frac{K[BIL]}{K[BLQ]}=\frac{BI\sin \frac{B}{2}}{BQ\sin \frac{A}{2}}=\frac{\sin C}{\sin \widehat{BIQ}}\cdot \frac{\sin \frac{B}{2}}{\sin \frac{A}{2}}\\ & =\frac{\sin C\sin \frac{B}{2}}{\sin \frac{A+B}{2}\sin \frac{A}{2}}=\frac{\sin C\sin \frac{B}{2}}{\cos \frac{C}{2}\sin \frac{A}{2}}=\frac{2\sin \frac{C}{2}\sin \frac{B}{2}}{\sin \frac{A}{2}},(2)\end{array}$$ since $$\widehat{BIQ}=180^{\circ }-\frac{\hat{B}}{2}-\widehat{ALB}=180^{\circ }-\frac{\hat{B}}{2}-\left(\hat{C}+\frac{\hat{A}}{2}\right)=\frac{\hat{A}+\hat{B}}{2}.$$ From (1) and (2) it follows $\frac{AI}{IQ}=\frac{IL}{LQ}$, hence $AI\cdot LQ=IL\cdot IQ$.