, 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 $\widehat{QBI} = \widehat{QIB} = 90^\circ - \frac{1}{2}\widehat{C}$.
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.

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