YAY I can solve middle level geometry! So barycentric coordinates work well here, and I will leave the details to the reader. Denote $M_a$ as the midpoint of $BC$ and $N$ as the midpoint of $CL$. A well-known lemma states that $Q \in AM_a$. Now use barycentric coordinates wrt $\triangle ABC$. So easily get $D(0:s-c:s-b)$, $E(s-c:0:s-a)$, $F(s-b:s-a:0)$. Now line $BI$ is $cx=az$, line $CI$ is $bx=ay$, line $AM_a$ is $y=z$, and line $EF$ is $(b+c-a)x-(c+a-b)y-(a+b-c)z=0$. Solving for $K, L, M, Q$, (using $Q=EF \cap AM_a$) we get $$K(a:c-a:a) \text{ and } L(a:b:b-a) \text{ and } Q(2a:b+c-a:b+c-a) \text{ and } M(0:a+b-c:b-c-a)$$Now easily get $N(a:b:3b-a)$. Solving for the line of $MN$, we get $$(a^2-2b^2-3ab-ac+2bc)x+a(c+a-b)y+a(a+b-c)z=0$$and line $CK$ is just $(c-a)x=ay$. Solving for $P$ gives us $P(a:c-a:2b+c)$. Now use the length formula $l^2=-a^2yz-b^2zx-c^2xy$, where $x,y,z$ are components of the displacement vector. Calculations show that $$PQ^2=\frac{ab(a+b-c)(a+b+c)}{4(b+c)^2}$$$$KQ^2 = \frac{a^2b(a+b-c)(a-b+c)}{4c(b+c)^2}$$$$AB^2=c^2$$$$BI^2 = \frac{ac(a+c-b)}{a+b+c}$$and now $$\frac{PQ^2}{KQ^2} = \frac{AB^2}{BI^2} = \frac{c(a+b+c)}{a(a+c-b)}$$which gives us $PQ^2 = \frac{AB^2 \cdot KQ^2}{BI^2}$, or $PQ=\frac{AB \cdot KQ}{BI}$ as required. $\blacksquare$

Using harmonic division and harmonic pencil. It will be more easier.

Here is my solution. In order to prove $PQ= \frac{AB \cdot KQ}{BI}$, we will prove that $\triangle KQP \sim \triangle IBA$. We have the following well-known lemma. Lemma. $\angle BLC =\angle BKC=90^{\circ}$ and $BLKC$ is cyclic. We have $(MD,BC)=-1$ so $2= (CM,BD) = (LM,KQ)$ implies $\frac{KL}{2KM}= \frac{QL}{QM}$ or $2KM \cdot QL= KL \cdot MQ$. On the other hand, we have $(KL,QM)= (BC,DM)=-1$ so $QK \cdot ML= QL \cdot MK$. From these two, we obtain that $KL \cdot QM= 2KQ \cdot ML$ or $\frac{KL}{2KQ} = \frac{ML}{MQ}$. From $Q$ draw a parrallel line to $LC$ and intersects $KC$ at $P'$. $P'M$ intersects $LC$ at $T'$ then we will have $$\frac{LT'}{QP'}= \frac{ML}{MQ}=\frac 12 \cdot \frac{KL}{KQ}= \frac 12 \cdot \frac{LC}{QP'}.$$This follows $LC=2LT'$ or $T'$ is the midpoint of $LC$ so $P \equiv P'$. Hence, $QP \parallel LC$. Thus, $\angle KQP= \angle KLC= \angle KBC$ (since $BLKC$ is cyclic). On the other hand, we also have $$\angle LKC = 90^{\circ}+ \angle LKB=90^{\circ}+ \tfrac 12 \angle C = \angle AIB.$$Thus, we obtain $\triangle AIB \sim \triangle PKQ$ or $PQ= \frac{AB \cdot KQ}{BI}$. $\blacksquare$

Let $N$ be the midpoint of segment $CL$, and let $PQ$ meet $CL$ at $E$. Then $(L,C;N,E)=P(L,K;M,Q)=I(C,B;M,D)=-1$, but $N$ is the midpoint of $CL$ so $E$ is at the point of infinity$\implies PQ\parallel CL$. It is well-known that $B,K,L,C$ lie on the circle with diameter $BC$, check that $\angle CLK=\angle CBK=\angle IBA$ and $\angle CKL= 90^{\circ}+ \frac{1}{2} \angle C=\angle AIB\implies \triangle IAB\sim \triangle KCL\sim \triangle KPQ$ from which the length relation follows. Remark: using the same method the result is still true even if the points $Q$ and $M$ are switched.

Let $BL$ and $CP$ intersect $S$ It is well know that $I$ is the orthocenter of the $\triangle BCS$$\implies$ $(C,B;D,M)=(K,L;Q,M)=P(C,L;R,Q)=-1$ If we intersect the last bundle with line $CL$ we see that intersection of $PQ$,$CL$ is in infinity thus they are parallel. $CKLB$ is cyclic (well known) thus $\angle KPQ=\angle KCL=\angle KBL=\frac{\alpha}{2}$,$\angle CKL=\frac{\pi}{2}+\frac{\gamma}{2}$ thus $\triangle AIB$ and $\triangle PQK$ are similar and we are done;

Let $R$ be on $CK$ with $MR\parallel CI$. Project $(B,C;D,M)$ through $I$ onto $EF$ to get $(K,L;Q,K)=-1$. Project $(C,L;N,P_\infty)$ through $M$ onto $CK$ to get $(C,K;P,R)=-1$. Because $CL$ and $MR$ are parallel $PQ$ is also parallel to those two lines. Hence, $KPQ$ is similar to $KCL$ and furthermore $\angle KLC=\angle KBC=\angle IBA$ and $\angle CKL=90^\circ+\angle LKI=90^\circ+\angle ICB=\angle AIB$. This means that $KPQ$ is similar to $IAB$ from which the result immediately follows.

$\angle EKI=\angle FKB=90^{\circ}-\tfrac{\angle A}{2}-\tfrac{\angle B}{2}=\tfrac{\angle C}{2}=\angle ECI\rightarrow ICKE$ is cyclic$\rightarrow \angle BKC=90^{\circ}$ $\angle LCK=(90^{\circ}-\tfrac{\angle B}{2})-\tfrac{\angle C}{2}=\tfrac{\angle A}{2}=\angle BAI$ and $\angle FKC=90^{\circ}+\tfrac{\angle C}{2}=\angle AIB\rightarrow \bigtriangleup CKL\sim \bigtriangleup AIB$ From Menelaus on $\bigtriangleup CKL$ and line $PT$: $\frac{ML}{MK}=\frac{PC}{PK}$ Also, $-1=(BCDM)\stackrel{I}{=}(KLQM)\rightarrow \frac{ML}{MK}=\frac{QL}{QK}$ $\rightarrow \frac{PC}{PK}=\frac{QL}{QK}\rightarrow PQ\parallel CL\rightarrow \bigtriangleup CKL\sim \bigtriangleup PKQ\rightarrow \bigtriangleup PQK\sim \bigtriangleup AIB\rightarrow PQ=\frac{AB \cdot KQ}{BI}$

It was not particularily hard this time (FKMO, this time)

Indeed, I solved 4 out of 6 and I'm hoping for top 50....

WLOG $\angle B \ge \angle C$. A simple angle chase shows that $BCKL$ is cyclic. Indeed, $\angle KLC = \angle KMC + \angle LCM = \frac{B}{2}-\frac{C}{2} + \frac{C}{2} = \frac{B}{2} =\angle KBC$. Let $H(W, X; Y, Z)$ denote that $W, X, Y, Z$ form a harmonic pencil. Then its well known that $H(M, D; B, C)$. By a perspectivity at $I$, $H(M, Q; K, L)$. Let $PQ$ hit $CL$ at $R$. We claim $R = P_{\infty}$. Indeed, by a perspectivity at $P$, $H(G, R; C, L)$. But since $G$ is the midpoint of $CL$, it follows $R = P_{\infty}$. Thus $PQ\parallel CL$, so $\angle QPK = \angle LCK = \frac{A}{2} = \angle BAI$ and $\angle QKP = \angle LKC = \angle BIA$. Thus $\triangle KPQ\sim\triangle IAB$ so $\frac{PQ}{KQ} = \frac{AB}{IB}$ as desired.

we ve proved before that $ BCKL$ are inscribed in the circle of diameter $BC$ then $\widehat{LKC}=\pi -\widehat{LBC}=\pi-(\hat A+\hat B )=\widehat{BIA}$ thus the desired result is equivalent to $BIA \sim PKQ$ . $(B,C;D,M)=-1 \implies (K,L;Q,M)=1\implies \frac{ML}{MK}= \frac{QL}{QK}$ besides applying menelaus with the triangle $LKC$ yields $\frac{ML}{MK}= \frac{PC}{PK}$ so $\frac{PC}{PK}= \frac{QL}{QK}$ then $PQ \parallel CL$ therefore $ \widehat{KQP}=\widehat{KLC}=\frac{\hat B}{2}$ then $BIA \sim PKQ$ . R HAS

Very nice problem. Here's my trigonometric approach. My solution: Let $P$ be point on $CK$ such that $PQ=\frac{AB\cdot KQ}{BI}$. From this we get $\triangle PKQ \sim \triangle AIB$. From the well know fact that $BFLK$ and $BLKC$ are cyclic we get $\angle BLC=\angle BKC=90$ That implies $KC=BC\sin{\frac{b}{2}}$ and $KQ=\frac{IK}{\sin{\frac{a}{2}+c}}\sin{90-\frac{b}{2}}= \frac{IC\sin{\frac{a}{2}}}{\sin{\frac{a}{2}+c}}\sin{90-\frac{b}{2}}=\frac{\frac{BC}{\sin{\frac{b+c}{2}}}\sin{\frac{b}{2}}\sin{\frac{a}{2}}}{\sin{\frac{a}{2}+c}}\sin{90-\frac{b}{2}}$// which implies that $KP=\frac{AI}{BI}KQ=\frac{\sin{\frac{b}{2}}}{\sin{\frac{a}{2}}}KQ= \frac{\sin{\frac{b}{2}}}{\sin{\frac{a}{2}}}\frac{\frac{BC}{\sin{\frac{b+c}{2}}}\sin{\frac{b}{2}}\sin{\frac{a}{2}}}{\sin{\frac{a}{2}+c}}\sin{90-\frac{b}{2}}=BC\frac{\sin{\frac{b}{2}}^2}{\cos{\frac{a}{2}}\sin{\frac{a+2c}{2}}}\cos{\frac{b}{2}}$ So we have that $PC=KC-KP=BC\sin{\frac{b}{2}}(1-\frac{\sin{\frac{b}{2}}}{\cos{\frac{a}{2}}\sin{\frac{a+2c}{2}}}\cos{\frac{b}{2}})=BC\sin{\frac{b}{2}}(\frac{ \cos{\frac{a}{2}}\sin{\frac{a+2c}{2}}-\frac{\sin{b}}{2}}{\cos{\frac{a}{2}}\sin{\frac{a+2c}{2}}}).$ Now by the Menelaus theorem on $\triangle KCL$ we get that the problem is equivalent to proving $\frac{PK}{CP}=\frac{KM}{ML}$ . Which is equivalent to $\frac{\frac{\sin{b}}{2}}{\cos{\frac{a}{2}}\sin{\frac{a+2c}{2}}-\frac{\sin{b}}{2}}=\frac{\sin{b}}{\sin{c}}$ which is equivalent to $\sin{c}=2\cos{\frac{a}{2}}\sin{\frac{a+2c}{2}}-\sin{b}$ which is true since $\sin{c}+\sin{b}=\sin{c}+\sin{a+c}=2\sin{\frac{a+2c}{2}}\cos{\frac{a}{2}}$. Therefore we have proven that this point $P$ is on line connecting $M$ and midpoint of $CL$ so the problem is finished. $\blacksquare$

First note that $(B,C;D,M) = -1$. By a well-known lemma $BLKC$ is cyclic with $\angle BLC = \angle BKC = 90^{\circ}$. Then by Brokard, if $BL \cap CK = Y$ and $X$ is the midpoint of $BC$ we have $MX \equiv BC \perp IY$. We also know that $ID \perp BC$ so $Y,Q,I,D$ are collinear. Now by taking the perspective of $(B,C;D,M)$ at $Y$ with respect to line $EF$ we also get that $(L,K;Q,M)=-1$. Therefore $P(L,K;Q,M)$ is an harmonic pencil meaning $PL,PC,PQ,PS$ is an harmonic pencil and we know that $S$ is the midpint of $LC$ so $QP \parallel FC$. Therefore $\triangle KQP \sim \triangle KLC$ and since $\angle KLC = \angle KBC = \angle IBA$ and $\angle LKC = 180^{\circ} - \angle LBC = 90^{\circ} + \angle LCB = 90^{\circ} + \angle \dfrac{C}{2} = \angle ABI$ so $\triangle KLC \sim \triangle IBA$. This gives $\triangle KQP \sim \triangle IBA$ which gives the desired $PQ = \dfrac{AB \cdot KQ}{BI}.$

Let $X$ be the midpoint of $CL$, and let $Y$ be on $CK$ such that $MY\parallel CL$. By Iran lemma, $(BCLK)$ is cyclic with diameter $BC$. The key claim is $PQ\parallel CL$. Indeed, \[(MQ;LK)\stackrel{I}{=}(MD;CB)=-1\]and \[(YP;CK)\stackrel{M}{=}(\infty X;CL)=-1,\]and since $CL\parallel YM$, we must then have $PQ\parallel CL$ as well, as desired. Now, we have \[\frac{KQ}{PQ}=\frac{KL}{CL}=\frac{\sin\angle LBK}{\sin\angle LBC}=\frac{\sin(A/2)}{\cos(C/2)}\]and \[\frac{BI}{AB}=\frac{\sin\angle BAI}{\sin\angle AIB}=\frac{\sin(A/2)}{\cos(C/2)},\]so $\frac{KQ}{PQ}=\frac{BI}{AB}$, as desired.

Claim: $PQ\parallel CL$. Proof: This is just harmonic chasing. Let $N$ be the midpoint of $CL$. Let $Q'\in EF$ such that $PQ' \parallel CL$. Then \[ -1 = (CL; N\infty_{CL}) \stackrel{P}{=} (KL; MQ') \stackrel{I}{=} (B,C;M,IQ'\cap BC). \]But $(BC;MD)=-1$ by Ceva-Menelaus, so $IQ'\cap BC=D$, and hence $Q'=ID\cap EF=Q$. $\square$ We want to show $\tfrac{PQ}{KQ}=\tfrac{AB}{BI}$. We claim that $\triangle AIB\sim \triangle PKQ$; this will imply the desired length ratio. We will prove this by AA similarity. We will show $\angle QKP=\angle AIB$. We know $\angle QKP=\angle MKC=\angle BKC+\angle MKB=90+\angle MKB$ since $\angle BKC=90$ by the Iran Lemma. And $\angle MKB=\angle (EF,BI)$ can be computed to be $C/2$. Hence $\angle QKP=90+C/2=\angle AIB$. We will show $\angle KPQ = \angle BAI$. Note that $BCKL$ is cyclic by the Iran Lemma. We have \begin{align*} \angle KPQ &= \angle KCL = \angle KCI = \angle KCB-\angle ICB \\ &= (90-\angle KBC)-C/2 = 90-B/2-C/2 = A/2=\angle BAI, \end{align*}as desired.

When you see such equation the first thing you wonder is whether $\triangle KQP \sim \triangle IBA$ and I will prove it. Let $x,y,z$ be complex numbers lying on unit circle centered at zero (coordinate of incenter), such that they represent points $D,E,F$ respectively on Argand plane. By already existing formulas $$b=\frac{2xz}{x+z}, \ c=\frac{2xy}{x+y},\ a=\frac{2yz}{y+z},\ m=\frac{x^2(y+z)-2xyz}{x^2-yz}.$$$$K\in BI\iff \frac{b}{k}=\overline{\left(\frac{b}{k}\right)}\iff\overline{k}=\frac{k}{xz},\ K\in EF\iff \overline{k}=\frac{y+z-k}{yz}$$Hence we get coordinates of $K$ and by swapping variables $y,z$ a formula for $l$ too $$k=\frac{x(y+z)}{x+y},\ l=\frac{x(y+z)}{x+z}.$$$$Q\in DI\iff \frac{q}{x}=\overline{\left(\frac{q}{x}\right)}\iff\overline{q}=\frac{q}{x^2},\ Q\in EF\iff \overline{q}=\frac{y+z-q}{yz}$$Therefore $$q=\frac{x^2(y+z)}{x^2+yz}$$Denote $N$ as the midpoint of $CL$ $$n=\frac{c+l}{2}=x\cdot\frac{y^2+3y(x+z)+xz}{(x+y)(x+z)}.$$The time's up for $P$. $$P\in CK\iff \frac{c-p}{c-k}=\overline{\left(\frac{c-p}{c-k}\right)}\iff \frac{2xy-p(x+y)}{x(y-z)}=\frac{2z-\overline{p}(x+y)z}{z-y}\iff \overline{p}=\frac{2x(y+z)-p(x+y)}{xz(x+y)}.$$Let's have an auxiliary calculation:$$n-m= \frac{2x(y-z)(x - y) (x^2 + 2 x z + y z) }{(x^2-yz)(x+y)(x+z)}.$$Hence$$\frac{m-p}{n-m}=\frac{x^2(y+z)-2xyz+(yz-x^2)p}{2x(y-z)(x-y)}\cdot \frac{(x+y)(x+z)}{x^2+2xz+yz},\ \overline{\left(\frac{m-p}{n-m}\right)} =\frac{(y+z-2x)xz(x+y)+(x^2-yz)\left(2x(y+z)-p(x+y)\right)}{2xz(y-z)(x-y)}\cdot \frac{y(x+z)}{x^2+2xy+yz}$$En route to equation on the right there is division by $x+y$ on nominator and denominator (just so you can have full factorization if you're still checking ). We have $$P\in MN\iff \frac{m-p}{n-m}= \overline{\left(\frac{m-p}{n-m}\right)}\iff$$It's just a linear equation in $p$. Strike a pose and let's get to it.$$p=\frac{x}{(x+y)(x^2-yz)}\cdot\frac{(x^2+2xz+yz)y(z(x+y)(y+z-2x)+2(y+z)(x^2-yz))-z(xy+xz-2yz)(x+y)(x^2+2xy+yz)}{y(x^2+2xz+yz)-z(x^2+2xy+yz)}$$$$p=\frac{x(y-z)(x^2-yz)\left(x^2(2y+z)+y^2z\right)}{(x+y)(y-z)(x^2-yz)(x^2+yz)}=\frac{x\left(x^2(2y+z)+y^2z\right)}{(x+y)(x^2+yz)}$$Thus $${p-q}=\frac{xy(x-y)(x-z)}{(x+y)(x^2+yz)}.$$We have also $$k-q=\frac{xy(y+z)(z-x)}{(x+y)(x^2+yz)}$$and $$\frac{p-q}{k-q}=\frac{y-x}{y+z}\implies \overline{\left(\frac{p-q}{k-q}\right)}=\frac{(x-y)z}{(y+z)x}.$$Since $$\frac{a-b}{0-b}=\frac{(x-y)z}{(y+z)x}=\overline{\left(\frac{p-q}{k-q}\right)}$$we have $\triangle KQP \sim \triangle IBA$ in different orientation.QED

A routine trigonometric exercise. [asy][asy] size(10cm); real labelscalefactor = 0.25; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.787015230130102, xmax = 7.30449740129611, ymin = -15.112225013524949, ymax = 3.6536835658586106; /* image dimensions */pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen ffvvqq = rgb(1,0.3333333333333333,0); pen zzttff = rgb(0.6,0.2,1); pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */draw(circle((-2.820378344729934,-4.273401634601451), 2.9668696812399857), linewidth(1.2) + blue); draw((-5.57372495950885,1.2996639065066862)--(-5.903860088690224,-7.228281975225372), linewidth(0.8) + fuqqzz); draw((-5.903860088690224,-7.228281975225372)--(6.584304079857037,-7.27693022203824), linewidth(0.8) + fuqqzz); draw((6.584304079857037,-7.27693022203824)--(-5.57372495950885,1.2996639065066862), linewidth(0.8) + fuqqzz); draw((-11.950642167279186,-7.204726443522506)--(0.5817647546063381,-1.0131502061929452), linewidth(0.8) + ffvvqq); draw((-11.950642167279186,-7.204726443522506)--(-5.903860088690224,-7.228281975225372), linewidth(0.8)); draw((-2.8142137004205567,-2.6909172652810813)--(-2.8319358412066746,-7.240248804533912), linewidth(0.8)); draw((-5.903860088690224,-7.228281975225372)--(0.5817647546063381,-1.0131502061929452), linewidth(0.8) + zzttff); draw((-4.762127032196102,-3.653274613919133)--(6.584304079857037,-7.27693022203824), linewidth(0.8) + zzttff); draw((-11.950642167279186,-7.204726443522506)--(4.396305736582375,-4.993706480437174), linewidth(0.8)); draw((0.5817647546063381,-1.0131502061929452)--(6.584304079857037,-7.27693022203824), linewidth(0.8)); draw((-2.8142137004205567,-2.6909172652810813)--(4.396305736582375,-4.993706480437174), linewidth(0.8)); draw(circle((0.34022199558340654,-7.252606098631803), 6.244129461992793), linewidth(0.8) + linetype("4 4") + zzttff); draw(circle((1.881962867563551,-5.775165928319848), 4.936325442140265), linewidth(0.8) + linetype("4 4") + qqwuqq); draw(circle((-4.36211921671008,-5.7508418049134145), 2.135367503074246), linewidth(0.8) + linetype("4 4") + qqwuqq); /* dots and labels */dot((-5.57372495950885,1.2996639065066862),dotstyle); label("$A$", (-5.225903580507468,1.4762047222592527), NE * labelscalefactor); dot((-5.903860088690224,-7.228281975225372),dotstyle); label("$B$", (-6.532390886667082,-7.985930616290686), NE * labelscalefactor); dot((6.584304079857037,-7.27693022203824),dotstyle); label("$C$", (6.908592157005317,-7.708796945287132), NE * labelscalefactor); dot((-2.8203783447299346,-4.273401634601452),linewidth(4pt) + dotstyle); label("$I$", (-3.3849441945552834,-4.521759728746252), NE * labelscalefactor); dot((-2.8319358412066746,-7.240248804533912),linewidth(4pt) + dotstyle); label("$D$", (-3.147401047980808,-8.164087976221543), NE * labelscalefactor); dot((-1.110172379314469,-1.8490437275839229),linewidth(4pt) + dotstyle); label("$E$", (-1.8607090040357328,-1.9681709030706411), NE * labelscalefactor); dot((-5.785027400677454,-4.158633692515978),linewidth(4pt) + dotstyle); label("$F$", (-6.354233526736226,-4.066468697811841), NE * labelscalefactor); dot((-4.762127032196102,-3.653274613919133),linewidth(4pt) + dotstyle); label("$L$", (-5.0675414827911505,-3.4528155691611127), NE * labelscalefactor); dot((0.5817647546063381,-1.0131502061929452),linewidth(4pt) + dotstyle); label("$K$", (0.415746150636323,-0.7012741213401055), NE * labelscalefactor); dot((-11.950642167279186,-7.204726443522506),linewidth(4pt) + dotstyle); label("$M$", (-12.470969551028967,-7.76818273193075), NE * labelscalefactor); dot((-2.8142137004205567,-2.6909172652810813),linewidth(4pt) + dotstyle); label("$Q$", (-2.731700541475476,-2.54223350729229), NE * labelscalefactor); dot((0.9110885238304673,-5.465102417978686),linewidth(4pt) + dotstyle); label("$N$", (0.673084559425338,-5.986609132622185), NE * labelscalefactor); dot((4.396305736582375,-4.993706480437174),linewidth(4pt) + dotstyle); label("$P$", (4.770703837835039,-4.6999170886771084), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Firstly, let's do some angle chasing. We have $\angle LIB=90^{\circ}-\frac{A}{2}=\angle AFL$ and $\angle IFB=\angle IDB=90^{\circ}$, hence $F,L,I,B,D$ are concyclic. Similarly $D,C,K,E,I$ are concyclic. Hence $\angle LKB=\angle ECI=\angle LCB$, which implies $L,K,C,B$ are concyclic. Secondly, we will use trigonometry. CLAIM. $QP\|LC$ Proof. We have $$\frac{KQ}{QL}=\frac{KQ}{QI}\cdot\frac{QI}{QL}=\frac{\sin\frac{B}{2}\cos\frac{B}{2}}{\sin\frac{C}{2}\cos\frac{C}{2}}$$On the other hand $$\frac{\sin\angle KMN}{\sin\angle CMN}=\frac{\sin\angle MLN}{\sin\angle LCM}=\frac{\sin\frac{B}{2}}{\sin\frac{C}{2}}$$Hence $$\frac{KP}{PC}=\frac{\sin\angle KMP}{\sin\angle CMP}\frac{\sin\angle MCK}{\sin\angle MKC}=\frac{\sin\frac{B}{2}\cos\frac{B}{2}}{\sin\frac{C}{2}\cos\frac{C}{2}}$$as desired. $\blacksquare$ Hence $\frac{KQ}{QP}=\frac{KL}{LC}=\frac{\sin\frac{A}{2}}{\cos\frac{C}{2}}=\frac{BI}{AB}$ as desired.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/11.%202.%20Parallele%20a%20une%20cevienne%20du%20triangle.pdf p. 6... Sincerely Jean-Louis