Lemma $ABC$ be a triangle, C excircle be $C$ and tangency points be $A', B', C'$ with centre $I_C$, radius $r$. The C median cuts $A'B'$ at $P$. Incentral triangle of $ABC$ be $A_1B_1C_1$ and orthic triangle be $A_2B_2C_2$. Then, $PC_1 \parallel CC'$, so that $\dfrac{r}{CC_2} = \dfrac{I_CP}{I_CC'}$ Proof Let the intouch triangle be $A_3B_3C_3$. Note $C_3I \cap \odot A_3B_3C_3 \in CC'$, so it follows $CC' \parallel IM$, where $M$ is the midpoint of $AB$ and $I$ is the incentre, as the gergonne and nagel points are isotomic conjugates. Now, also, $P$ is the pole of the parallel through $C$ to $AB$, so it lies on $I_CC'$. Hence, it suffice to show $\dfrac{MC'}{C'C_2} = \dfrac{C_1C_3}{C_1C_2}$. Now note $(I_C, I; C_1, C) = -1$ so dropping perpendiculars is follows $(C_2, C_1; C_3, C') = -1$. So, $(MC')^2 = MC_1 \cdot MC_2 \implies \dfrac{MC_2}{MC'} = \dfrac{MC'}{MC_1} \implies \dfrac{C'C_2}{MC_3} = \dfrac{C'C_1}{MC_1} = \dfrac{C_1C_2}{C_1C_3}$, so take inverse we get what we want. Main Proof We must show $BYC$ and $I_CC'D$ are perspective. This means that if $I_CC' \cap CY = T, \; YB \cap C'D = U, \; CB \cap I_CD = V$, then $T, U, V$ are collinear by Desagues theorem. Let $M_B$ be the midpoint of $AC$. The cross ratio $(AM_B, XC)$, by taking persepctive from $D$, is $\dfrac{BC}{BA'} = \dfrac{a}{s-a}$, so if $Z= BC \cap LD$ then: \[\dfrac{UC'}{UD} \cdot \dfrac{YD}{YC'} = \dfrac{UC'}{UD} \cdot \dfrac{c}{s-a} = \dfrac{a}{s-a} \implies \dfrac{UC'}{UD} = \dfrac{a}{c}\] Now, height from $D$ to $BC$ is equal to height from $A$ to $BC$ so if $r$ is the C-exradius, then $\dfrac{DV}{DI_C} = \dfrac{bc}{2Rr}$ where $R$ is the radius of $\odot ABC$. By the lemma, we have $\dfrac{I_CP}{I_CC'} = \dfrac{2Rr}{ab}$ so: \[\dfrac{UC'}{UD} \cdot \dfrac{I_CP}{I_CC'} \cdot \dfrac{I_CP}{I_CC'} = \dfrac{a}{c} \cdot \dfrac{2Rr}{ab} \cdot \dfrac{bc}{2Rr} = 1\] Therefore $TUV$ is a line by Menelaus theorem.

Denote $BC=a,$ $CA=b,$ $AB=c,$ $s=\tfrac{1}{2}(a+b+c)$ (b>a). Let $CC',CC_1,CL$ cut $AD$ at $E,F,G,$ resp. Clearly $AE=b$ and $AF=a$ $\Longrightarrow$ $FE=b-a$ and $\triangle LBC \sim \triangle LAG$ gives $\frac{GA}{a}=\frac{s-b}{s-a} \Longrightarrow FG=a-a \cdot \frac{s-b}{s-a}=\frac{a(b-a)}{s-a} \Longrightarrow$ $\frac{FE}{FG}=(b-a) \cdot \frac{s-a}{a(b-a)}=\frac{s-a}{a}=\frac{BK}{BC}$ $\Longrightarrow$ $D(C,B,K,A)=C(G,F,E,B)$ $\Longrightarrow$ $D(C,B_1,X,A)=B(C,B_1,X,A)=C(B,C',C_1,L)$ $\Longrightarrow$ $CC' \cap BB_1,$ $Y \equiv BX \cap CC_1$ and $L$ are collinear, i.e. $YL,BB_1$ and $CC'$ concur.

We solve the problem using barycentric coordinates: We have $A = (1,0,0)$; $B = (0,1,0)$; $C = (0,0,1)$. Since $ABCD$ is a parallelogram, we get \[D = A + C - B = (1,-1,1)\] Because $BK = BL = p - a$, we have \[K = (0:p:a-p) \\ L = (p-a:p-b:0)\] In order to find out the coordinates of $X$, we have to deduce the equation of the line $DK$: \[ D \in DK \Rightarrow u - v + w = 0 \ (1) \\ K \in DK \Rightarrow vp + w(a-p) = 0 \ (2)\] From $(1)$ and $(2)$ the equation follows: \[ ax + (a-p)y - pz = 0\] Now, since the equation of the $AC$ is obviously $y = 0$, the coordinates of $X$ are: \[X = (p:0:a)\] Looking for the coordinates of $Y$, we write the equations of the lines $XB$ and $CC_1$: \[ XB : \frac{x}{z} = \frac{p}{a}\ (3) \\ CC_1 : x = y\ (4)\] From $(3)$ and $(4)$, we obtain the coordinates of $Y$: \[ Y = (p:p:a) \] Let $\{T\} = \text{bis}\ (\angle ACB) \cap BB_1$. We must show that $T \in LY$. Since the equation of $\text{bis}\ (\angle ACB)$ is $\frac{x}{y} = \frac{a}{b}$, and the equation of $BB_1$ is $x=z$, we get that \[ T = (a:b:a) \] In order to find out the equation of $LY$, we must solve the system \[ Y \in LY \Rightarrow up + vp + wa\ (5) \\ L \in LY \Rightarrow u(p-a)+v(p-b)=0\ (6) \] After some computations, one can find: \[ LY : a(p-b)x + a(a-p)y + p(b-a)z = 0\] Plugging in the coordinates of $T$, we finally get, after some other calculations, that $T \in LY$, namely the three lines are concurrent. $\square$