Nice problem. Notice that the configuration is the one in the IMO Shortlist 2005 G1 problem. That one was very nice too. Proof: Note that the inverse of $ K$ in $ (I)$ is the intersection of $ BI$ and the perpendicular from $ A$ onto $ BI$. This point lies on the line connecting the midpoints of $ AB$ and $ AC$. For a proof, see *. Similar arguments for $ L$ show that the inverse of the circumcircle of $ ILK$ is the line connecting the midpoints of $ AB$ and $ AC$. This line is tangent to $ (I)$ if and only if $ h_A = 4r$. Hence the condition is equivalent to $ \left|BC\right|\cdot 2r = S = s\cdot r$. We find that the condition is equivalent to $ s = 2 \left|BC\right|$, from which the relation follows. Remarks: Note that $ S,r,s,h_A$ denote the area, inradius, semiperimeter and heigth from $ A$ in triangle $ ABC$. For a proof of the starred property, you could easiley try this yourself, or look at Lemma 2 in: http://forumgeom.fau.edu/FG2008volume8/FG200827.pdf

What did you define as D and could you explain further why KIL is similar to BIC?

Also, does anyone know of a projective solution to this problem, if there is one?

Here's a quick and (probably) op way but no paper or pencil is needed. It's well-known that $BCKL$ is cyclic and $\angle BKC=90^{\circ}$ $\implies$ $CK\cap BL=\{H_A\}$ is the orthocenter of $\triangle IAB$.Invert thru $I$ with radius $\sqrt{IK\cdot IB}$ and compose it with central symmetry .$\odot IKL$ gets taken to $BC$ $\implies$ $\odot I$ stays fixed $\implies$ $IK\cdot IB=r^2=ID\cdot IH_A$ $\implies$ the orthocenter of $\triangle IBC$ is the antipode of $D$ in $\odot I$ but $H_A$ is also the pole of $A$-midline $\implies$ $A$-midline is tangent to $\odot I$ $\implies$ $h_a=2r$ and hence the desired.$\blacksquare$

I found a really beautiful solution to this problem. Here we go....First let the $(ILK)$ is tangent to $(I)$ at point $T$. So we have that the center of $(ILK)$ lies on $IT$. Therefore we get $\measuredangle ILT=90^{\circ}=\measuredangle BLC$( because of the cyclic quadrilateral $BPLI$) . Hence $BL \cap CK=T$ and moreover $T,I,R$ are collinear, where $R=(I) \cap BC$. Let $BT \cap (I)=D $ . From angle chasing we get that $\measuredangle DQT= \measuredangle RPQ$ , which means that $Q,I,D$ are collinear. Let $BD \cap AC=X$ .It is well known that the point $X$ is the point, where the $B-excircle$ meets $AC$. We get that $\triangle BCX $ is isosceles, which means $BC=CX$ . Now the conclusion follows immediately(the converse can be proved analogously).

k.vasilev wrote: From angle chasing we get that $\measuredangle DQT= \measuredangle RPQ$ We get that $$\angle RPQ = \angle RTQ = \angle ITQ = \angle IQT$$not $\angle RPQ = \angle DQT$. Please correct me if i am wrong

sinus law

Let $D$, $E$, $F$ be the tangency point of the incircle on $BC$, $CA$, $AB$ respectively. Let $Z=CK\cap BL$. Claim: Center of $(IKL)$ lies on $DI$. Proof. By Iran lemma we have $\angle BKC=\angle BLC=90^\circ$. So $IKZL$ is cyclic and center of $(IKZL)$ lies on $ZI$. Now by Brocard Theorem on cyclic quadrilateral $IKZL$, we have $ZI\perp BC$. Therefore we have $ZI\equiv ID$. $\square$ Claim: $(IKL)$ is tangent to $M_aM_b$, $M_aM_c$ at $K$, $L$. Proof. Notice that $\angle LKM_b=\angle M_bEK+\angle EM_bK=(90^\circ-\angle A/2)+\angle A=90^\circ+\angle A/2$. And it is known that $\angle LIK=90^\circ+\angle A/2$. $\square$ Claim: Line $M_bM_c$ is the radical axis of the incircle and $(IKL)$. Proof. Notice that $\triangle M_bEK\sim\triangle M_aKL$. Since $\triangle M_aKL$ is isosceles hence $\triangle M_bEK$ is also isosceles with $M_bE=M_bK$ and both are tangent to the incircle and $(IKL).$ $\square$ Now come to the conclusion. From all the 3 claims, we can say if the incircle and $(IKL)$ are tangent then the incircle $(DEF)$ would be the $A-$excircle of $\triangle AM_bM_c$. Now by \emph{Pitot Theorem} we can say \begin{align*} AB+AC=3BC&\iff a+\frac{a}{2}=\frac{b}{2}+\frac{c}{2} \\ &\iff \text{The incircle is inscribed in $BCM_bM_c$} \\ &\iff \text{$(DEF)$ and $(IKL)$ are tangent.} \end{align*}and we are done.