Korean Mathematical Olympiad

Since $BD$ and $BF$ are tangent lines to the incircle of $△ABC$, $BD=BF$. But the line $BI$ bisects $\angle DBF$, so $DF$ and $BI$ are perpendicular to each other. Similarly, $DE$ and $CI$ are perpendicular to each other. It follows that $\angle BKD=\angle BKF=90^{\circ }-\angle DFK$ and $\angle CED+\angle ECI=90^{\circ }$. Since $AC$ is tangent to

the excircle of $△DEF$, $\angle DFK=\angle CED$, and so $\angle BKD=90^{\circ }-\angle CED$. Thus $\angle BKD=\angle ECI=\angle DCI$, which means that the point $K$ lies on the excircle of the quadrilateral $CEID$. Therefore $\angle BKC=\angle IEC=90^{\circ }$. In a similar way, it can be proved that $\angle BLC=90^{\circ }$. Applying Menelaus theorem to $△DKL$ with respect to the line $MP$ leads to the equality $\frac{KP\cdot CJ}{JL\cdot MK}=1$, from which we get $\frac{KP}{PC}=\frac{MK}{LM}$, as $CJ=JL$. Since $DI$ and $DM$ bisect the internal angle and external angle of $△DKL$ at $D$, respectively, $\frac{KQ}{QL}=\frac{KD}{DL}=\frac{KM}{ML}$ holds. Combining this with $\frac{KP}{PC}=\frac{MK}{LM}$, we get $\frac{KP}{PC}=\frac{KQ}{QL}$. It follows that $PQ$ and $CL$ are parallel to each other. Let $A^{\prime }$ be the intersection point of $BL$ and $CK$. Note that $I$ is the orthocenter of the $△A^{\prime }BC$. It can easily be seen that $\angle B{A}^{\prime }D=\frac{1}{2}\angle BCA$ and $\angle C{A}^{\prime }D=\frac{1}{2}\angle ABC$, and so $\angle B{A}^{\prime }C=\frac{1}{2}(\angle ABC+\angle ACB)$. It follows that $\angle KPQ=\angle A^{\prime }CL=90^{\circ }-\angle B{A}^{\prime }C=\frac{1}{2}\angle BAC=\angle IAB$. Since the points $A^{\prime }$, $L$, $C$, $D$ are concyclic, $\angle A^{\prime }KL=\angle A^{\prime }BC$, from which we obtain $\angle PKQ=90^{\circ }+\frac{1}{2}\angle ACB=\angle AIB$. Since $\angle KPQ=\angle IAB$ and $\angle PKQ=\angle AIB$, $△KPQ$ and $△IAB$ are similar to each other, and so $$\frac{PQ}{QK}=\frac{AB}{B{I}^{\prime }}$$ from which the conclusion follows.