$ BD$ is the bisector of angle $ ABC$ and $ \angle EDF=\frac{1}{2}\angle ADC=90^o-\frac{1}{2}\angle EBF$ therefore $ D$ is the B-excenter of triangle $ EBF$. Thus $ L$ is the tangency of incenter $ (I)$ of triangle $ EBF$ and $ EF$. Let $ J=BD\cap EF$ $ \frac{r_b}{r}=\frac{DK}{IL}=\frac{KJ}{LJ}=\frac{ML}{LJ}+\frac{MJ}{LJ}=1+2\frac{MJ}{LJ}$ $ \frac{r_b}{r}=\frac{DJ}{IJ}=\frac{DB}{IB}$ then it easy to show that $ \frac{MJ}{LJ}=\frac{DJ}{BJ}$ $ \Rightarrow BL//DM$

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Let $ G \in AC$. $ CG=EA \Rightarrow \triangle EAD = \triangle GCD$, $ BEDG$ is cyclic, $ DF$ is a bisector of $ \angle EDF$. $ DF \cap (BEDG)=M$, $ BM \cap DE=N$, $ ME \cap BD=W$, $ FE \cap WD=S$, $ I$-midpoint of $ WD$. $ DM$ is a diameter of $ (BEDG) \Rightarrow ME$ and $ DB$ are altitudes of $ \triangle MND$ and $ BWEN$ is cyclic $ \Rightarrow \angle NDM=\angle NBE=\angle NWE$. $ \angle EWD=\angle DFG=\angle DFE \Rightarrow FWFD$ is cyclic $ \Rightarrow \angle EWF=180^0 - \angle NDM$. Then $ \angle NWE+\angle EWF=180^0$ and $ N, W, F$ lie on the same line and $ NF$ is an altitude. We shall prove: For an arbitrary quadriateral $ XYZT$ such that $ \angle Y=\angle T=90^0$, if $ X'$ and $ Z'$ are the projections of $ X$ and $ Z$ on $ YT$, then $ YX'=TZ'$. $ \triangle X'XY \sim \triangle Z'YZ$ $ \triangle XX'T \sim \triangle TZ'Z$ Then, $ \frac {XX'}{YX'}= \frac {YZ'}{ZZ'}$ and $ \frac {XX'}{TX'}= \frac {TZ'}{ZZ'} \Rightarrow \frac{TX'}{YX'}= \frac{YZ'}{TZ'} \Rightarrow YX'=TZ'$ Then $ WL \perp FE$ and $ \triangle SWL \sim \triangle SIM \Rightarrow \frac{SW}{SI}=\frac{SL}{SM}$. $ \angle FIE=2 \angle FDE$ $ \angle FBE=\angle FBW+\angle EBW=\angle EMD+\angle FND=180^0-2\angle FDE$. Then, $ BEIF$ is cyclic. We know that $ FWED$ is cyclic, so $ BS \cdot SI=FS \cdot SE=WS\cdot SD \Rightarrow \frac{SW}{SI}=\frac{BS}{SD} \Rightarrow\frac{SL}{SM}=\frac{BS}{SD}\Rightarrow \triangle SBL\sim\triangle SDM \Rightarrow BL \parallel MD$.

livetolove212 wrote: ................................ then it easy to show that $ \frac {MJ}{LJ} = \frac {DJ}{BJ}$ $ \Rightarrow BL//DM$ Can someone explain how we arrive to the first relation ? Babis

stergiu wrote: livetolove212 wrote: ................................ then it easy to show that $ \frac {MJ}{LJ} = \frac {DJ}{BJ}$ $ \Rightarrow BL//DM$ Can someone explain how we arrive to the first relation ? Babis $ 2\frac{MJ}{LJ}+1=\frac{DB}{IB}=\frac{EB+BF+EF}{EB+BF-EF}=\frac{2EF}{EB+BF-EF}+1$ $ =2\frac{1}{\frac{EB+BF-EF}{EF}}+1=2\frac{1}{\frac{EB+BF}{EF}-1}+1$ $ =2\frac{1}{\frac{BF}{\frac{BF.EF}{EB+BF}}-1}+1=2\frac{1}{\frac{BF}{FJ}-1}+1$ $ =2\frac{1}{\frac{BD}{DJ}-1}+1=2\frac{1}{\frac{BJ}{DJ}}+1=2\frac{DJ}{BJ}+1$

it's easy to prove that $D$ is the excenter of triangle $BEF$ hence the result is trivialized by ex-Boone theorem.