Let $A_2$, $B_2$, $C_2$ the projections of $F$, $E$, $D$ in $AC$, $BC$, $AB$ it's well-know $A_1A_2B_1B_2C_1C_2$ is cyclic and $ACC_1C_2$ is cyclic $\Longrightarrow$ $\measuredangle BAC=\measuredangle C_1C_2A_1=\measuredangle C_1B_1A_1$ similarly we get $\measuredangle ACB=\measuredangle C_1A_1B_1$ and $\measuredangle ABC=\measuredangle A_1C_1B_1$ hence $\triangle ABC\sim \triangle B_1A_1C_1$.

My idea is to calculate $A_{1}B_{1},C_{1}B_{1},A_{1}C_{1}$ with the angles and lengths of $ABC$ by cosines theorem I did it but I couldn't prove the similarity If some one have a complete solution please post Thanks

Drop other perpendiculars namely $C',A',B'$ from $D,E,F$ on $AB,BC,AC$.We have that $A_1A'B_1B'C_1C'$ are concyclic on the Taylor's circle but as $C_1B'||BC$ ($C_1B'$ is anti parallel to $FE$) we have : $\angle C_1A_1B_1=\pi-\angle C_1B'B_1=\gamma$ Analogously for other angles and hence the conclusion follows.

If we get $ X , Y , Z$ on $BC , AC , AB$ such that $ A_1Z||AC , B_1X||AB , C_1Y||BC $ , It's easy to see that $A_1ZC_1YB_1X$ are cyclic and the else is easy

hi. my solution very close to Arefe (Above) but i try to write this with more details: as above we let point $X,Y,Z$ such that: $ A_1Z||AC , B_1X||AB , C_1Y||BC $. NOW we claim that: CLAIM:$A_1ZC_1YB_1X$ cyclic. first we prove $C_1YB_1Z$, and $YB_1XA_1 $ Cyclic . From them we can conclude we are claim. if $C_1YB_1Z$ be cyclic should $FE||B_1Z \implies \frac{EB_1}{FZ}=\frac{AB}{AC}$ which implies from simple calculation . we can do this similarly for $YB_1XA_1 $. hence done. $\blacksquare$

Complete solution based on the law of cosines: First of all, note that throughout the solution, we will consider $\triangle{ABC}$ to be acute-angled. The other cases can be treated analogously. We will prove that $\triangle{ABC}$ is similar to $\triangle{B_1C_1A_1}$ in this order. More specifically, we will prove that $$\frac{AB}{B_1C_1}=\frac{BC}{C_1A_1}=\frac{CA}{A_1B_1}=\lambda$$Using the law of cosines in $\triangle{EB_1C_1}$, we obtain: $$B_1C_1^2=EB_1^2+EC_1^2-2EB_1*EC_1*cos(\angle{B_1EC_1})$$Note that $\angle{B_1EC_1}=90^\circ+\angle{A} \Longrightarrow cos(\angle{B_1EC_1})=-sin(\angle{A})$. Also, we have: $$\frac{EB_1}{EC}=\frac{BD}{BC} \Longleftrightarrow EB_1=\frac{BD*EC}{BC}=BD*cos(\angle{C})=AB*cos(\angle{B})*cos(\angle{C})$$$$\frac{EC_1}{FC}=\frac{AE}{AC} \Longleftrightarrow EC_1=\frac{AE*FC}{AC}=AE*sin(\angle{A})=AB*sin(\angle{A})*cos(\angle{A})$$By replacing, we find that: $$B_1C_1^2=(AB*cos(\angle{B})*cos(\angle{C}))^2+(AB*sin(\angle{A})*cos(\angle{A}))^2+2AB^2*sin^2(\angle{A})*cos(\angle{A})*cos(\angle{B})*cos(\angle{C}) \Longleftrightarrow$$$$(\frac{B_1C_1}{AB})^2=cos^2(\angle{B})*cos^2(\angle{C}))+sin^2(\angle{A})*cos^2(\angle{A}))+2sin^2(\angle{A})*cos(\angle{A})*cos(\angle{B})*cos(\angle{C})$$Similarly, one can prove that $$(\frac{C_1A_1}{BC})^2=cos^2(\angle{C})*cos^2(\angle{A}))+sin^2(\angle{B})*cos^2(\angle{B}))+2sin^2(\angle{B})*cos(\angle{A})*cos(\angle{B})*cos(\angle{C})$$$$(\frac{A_1B_1}{CA})^2=cos^2(\angle{A})*cos^2(\angle{B}))+sin^2(\angle{C})*cos^2(\angle{C}))+2sin^2(\angle{C})*cos(\angle{A})*cos(\angle{B})*cos(\angle{C})$$Therefore, it suffices to prove that: $$(\frac{B_1C_1}{AB})^2=(\frac{C_1A_1}{BC})^2 \Longleftrightarrow$$$$cos^2(\angle{B})*cos^2(\angle{C}))+sin^2(\angle{A})*cos^2(\angle{A}))+2sin^2(\angle{A})*cos(\angle{A})*cos(\angle{B})*cos(\angle{C})= cos^2(\angle{C})*cos^2(\angle{A}))+sin^2(\angle{B})*cos^2(\angle{B}))+2sin^2(\angle{B})*cos(\angle{A})*cos(\angle{B})*cos(\angle{C}) \Longleftrightarrow$$$$cos^2(\angle{C})(cos^2(\angle{B})-cos^2(\angle{A}))+2cos(\angle{A})*cos(\angle{B})*cos(\angle{C})(sin^2(\angle{A})-sin^2(\angle{B}))= sin^2(\angle{B})-sin^4(\angle{B})-sin^2(\angle{A})+sin^4(\angle{A}) \Longleftrightarrow$$$$cos^2(\angle{C})(sin^2(\angle{A})-sin^2(\angle{B}))+2cos(\angle{A})*cos(\angle{B})*cos(\angle{C})(sin^2(\angle{A})-sin^2(\angle{B}))= (sin^2(\angle{A})-sin^2(\angle{B}))(sin^2(\angle{A})+sin^2(\angle{B})-1) \Longleftrightarrow$$$$cos^2(\angle{C})+2cos(\angle{A})*cos(\angle{B})*cos(\angle{C})=sin^2(\angle{A})+sin^2(\angle{B})-1 \Longleftrightarrow$$$$2cos(\angle{A})*cos(\angle{B})*cos(\angle{C})=1-cos^2(\angle{A})-cos^2(\angle{B})-cos^2(\angle{C}) \Longleftrightarrow$$$$-2cos(\angle{A})*cos(\angle{B})*cos(\angle{A}+\angle{B})=1-cos^2(\angle{A})-cos^2(\angle{B})-cos^2(\angle{A}+\angle{B}) \Longleftrightarrow$$$$2cos(\angle{A})*cos(\angle{B})*(cos(\angle{A})*cos(\angle{B})-sin(\angle{A})*sin(\angle{B}))=cos^2(\angle{A})+cos^2(\angle{B})+(cos^2(\angle{A})*cos^2(\angle{B})-2sin(\angle{A})*sin(\angle{B})*cos(\angle{A})*cos(\angle{B})+sin^2(\angle{A})*sin^2(\angle{B}))-1 \Longleftrightarrow$$$$cos^2(\angle{A})*cos^2(\angle{B})=cos^2(\angle{A})+cos^2(\angle{B})+sin^2(\angle{A})*sin^2(\angle{B})-1 \Longleftrightarrow$$$$(1-sin^2(\angle{A}))*(1-sin^2(\angle{B}))=cos^2(\angle{A})+cos^2(\angle{B})+sin^2(\angle{A})*sin^2(\angle{B})-1 \Longleftrightarrow$$$$1-sin^2(\angle{A})-sin^2(\angle{B})=cos^2(\angle{A})+cos^2(\angle{B})-1 \Longleftrightarrow$$$$2=(sin^2(\angle{A})+cos^2(\angle{A}))+(sin^2(\angle{B})+cos^2(\angle{B}))$$which is true. For the calculations, we used some well-known trigonometric identities. The rest of the equalities can be proven in an analogous way. The solution is complete.

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Neo-Pythagorean wrote: Complete solution based on the law of cosines: First of all, note that throughout the solution, we will consider $\triangle{ABC}$ to be acute-angled. The other cases can be treated analogously. We will prove that $\triangle{ABC}$ is similar to $\triangle{B_1C_1A_1}$ in this order. More specifically, we will prove that $$\frac{AB}{B_1C_1}=\frac{BC}{C_1A_1}=\frac{CA}{A_1B_1}=\lambda$$Using the law of cosines in $\triangle{EB_1C_1}$, we obtain: $$B_1C_1^2=EB_1^2+EC_1^2-2EB_1*EC_1*cos(\angle{B_1EC_1})$$Note that $\angle{B_1EC_1}=90^\circ+\angle{A} \Longrightarrow cos(\angle{B_1EC_1})=-sin(\angle{A})$. Also, we have: $$\frac{EB_1}{EC}=\frac{BD}{BC} \Longleftrightarrow EB_1=\frac{BD*EC}{BC}=BD*cos(\angle{C})=AB*cos(\angle{B})*cos(\angle{C})$$$$\frac{EC_1}{FC}=\frac{AE}{AC} \Longleftrightarrow EC_1=\frac{AE*FC}{AC}=AE*sin(\angle{A})=AB*sin(\angle{A})*cos(\angle{A})$$By replacing, we find that: $$B_1C_1^2=(AB*cos(\angle{B})*cos(\angle{C}))^2+(AB*sin(\angle{A})*cos(\angle{A}))^2+2AB^2*sin^2(\angle{A})*cos(\angle{A})*cos(\angle{B})*cos(\angle{C}) \Longleftrightarrow$$$$(\frac{B_1C_1}{AB})^2=cos^2(\angle{B})*cos^2(\angle{C}))+sin^2(\angle{A})*cos^2(\angle{A}))+2sin^2(\angle{A})*cos(\angle{A})*cos(\angle{B})*cos(\angle{C})$$Similarly, one can prove that $$(\frac{C_1A_1}{BC})^2=cos^2(\angle{C})*cos^2(\angle{A}))+sin^2(\angle{B})*cos^2(\angle{B}))+2sin^2(\angle{B})*cos(\angle{A})*cos(\angle{B})*cos(\angle{C})$$$$(\frac{A_1B_1}{CA})^2=cos^2(\angle{A})*cos^2(\angle{B}))+sin^2(\angle{C})*cos^2(\angle{C}))+2sin^2(\angle{C})*cos(\angle{A})*cos(\angle{B})*cos(\angle{C})$$Therefore, it suffices to prove that: $$(\frac{B_1C_1}{AB})^2=(\frac{C_1A_1}{BC})^2 \Longleftrightarrow$$$$cos^2(\angle{B})*cos^2(\angle{C}))+sin^2(\angle{A})*cos^2(\angle{A}))+2sin^2(\angle{A})*cos(\angle{A})*cos(\angle{B})*cos(\angle{C})= cos^2(\angle{C})*cos^2(\angle{A}))+sin^2(\angle{B})*cos^2(\angle{B}))+2sin^2(\angle{B})*cos(\angle{A})*cos(\angle{B})*cos(\angle{C}) \Longleftrightarrow$$$$cos^2(\angle{C})(cos^2(\angle{B})-cos^2(\angle{A}))+2cos(\angle{A})*cos(\angle{B})*cos(\angle{C})(sin^2(\angle{A})-sin^2(\angle{B}))= sin^2(\angle{B})-sin^4(\angle{B})-sin^2(\angle{A})+sin^4(\angle{A}) \Longleftrightarrow$$$$cos^2(\angle{C})(sin^2(\angle{A})-sin^2(\angle{B}))+2cos(\angle{A})*cos(\angle{B})*cos(\angle{C})(sin^2(\angle{A})-sin^2(\angle{B}))= (sin^2(\angle{A})-sin^2(\angle{B}))(sin^2(\angle{A})+sin^2(\angle{B})-1) \Longleftrightarrow$$$$cos^2(\angle{C})+2cos(\angle{A})*cos(\angle{B})*cos(\angle{C})=sin^2(\angle{A})+sin^2(\angle{B})-1 \Longleftrightarrow$$$$2cos(\angle{A})*cos(\angle{B})*cos(\angle{C})=1-cos^2(\angle{A})-cos^2(\angle{B})-cos^2(\angle{C}) \Longleftrightarrow$$$$-2cos(\angle{A})*cos(\angle{B})*cos(\angle{A}+\angle{B})=1-cos^2(\angle{A})-cos^2(\angle{B})-cos^2(\angle{A}+\angle{B}) \Longleftrightarrow$$$$2cos(\angle{A})*cos(\angle{B})*(cos(\angle{A})*cos(\angle{B})-sin(\angle{A})*sin(\angle{B}))=cos^2(\angle{A})+cos^2(\angle{B})+(cos^2(\angle{A})*cos^2(\angle{B})-2sin(\angle{A})*sin(\angle{B})*cos(\angle{A})*cos(\angle{B})+sin^2(\angle{A})*sin^2(\angle{B}))-1 \Longleftrightarrow$$$$cos^2(\angle{A})*cos^2(\angle{B})=cos^2(\angle{A})+cos^2(\angle{B})+sin^2(\angle{A})*sin^2(\angle{B})-1 \Longleftrightarrow$$$$(1-sin^2(\angle{A}))*(1-sin^2(\angle{B}))=cos^2(\angle{A})+cos^2(\angle{B})+sin^2(\angle{A})*sin^2(\angle{B})-1 \Longleftrightarrow$$$$1-sin^2(\angle{A})-sin^2(\angle{B})=cos^2(\angle{A})+cos^2(\angle{B})-1 \Longleftrightarrow$$$$2=(sin^2(\angle{A})+cos^2(\angle{A}))+(sin^2(\angle{B})+cos^2(\angle{B}))$$which is true. For the calculations, we used some well-known trigonometric identities. The rest of the equalities can be proven in an analogous way. The solution is complete. Who hurt you?

Let $DY$ be perpendicular to $AB$, $FX$ be perpendicular to $AC$ and $EZ$ be perpendicular to $BC$. Claim1 : $XB_1ZA_1YC_1$ is cyclic. Proof : Note that $XEFC_1$ is cyclic and $EF || B_1Y$ so $XB_1YC_1$ is cyclic. with same approach $YC_1ZA_1$ and $ZA_1XB_1$ are cyclic so $XB_1ZA_1YC_1$ is cyclic. $\angle{ACB} = \angle{B_1DA} = \angle{B_1YA} = \angle{B_1A_1C_1}$ , $\angle{BAC} = \angle{C_1DB} = \angle{C_1B_1A_1}$ and $\angle{ABC} = \angle{A_1C_1B_1}$. we're Done.

$\angle{B_1C_1A_1} = C \Longleftrightarrow B_1C_1$ be tangent to circumcircle of $A_1CC_1$ and it means $\angle{DA_1C_1} = \angle{B_1C_1F}$ by key lemma in triangle $A_1CD$ : $\frac{Sin(DA1C1)}{Sin(90+DA1C1)} = f(DA1C1)$ d = C1D/CC1 . A1C/A1D = C1D/CC1.tgC C1D = AF.BC1/FB = AFCosB = bCosACosB CC1 = FCSinB = a(SinB)^2 => f(DA1C1) = bCosA.CosB / aSinB.SinB.tgC = 1/tgA.tgB.tgC => f(DA1C1)=f(B1C1F) => DA1C1 = B1C1F => B1C1A1 = C B1A1C1 = A C1B1A1 = B

It is easy to show that: $CA_{1}=sinA sinBb , B_{1}C=cos^2Cb$.By cosine law in $CB_{1}A_{1}$ triangle we'd have $(B_{1}A_{1})^2$=$b^2(cos^4C+sin^2Asin^2B-2sinA sinB cos^3C)$.By trigonometric identities it's not hard to show $B_{1}A_{1} =\sqrt{(cosA cosB cosC)^2+(sinA sinB sinC)^2}b$.so $A_{1}B_{1}$ is a multiple of b which is symetric with respect of A,B and C. Similarly $B_{1}C_{1}$ and $C_{1}A_{1}$ have the same ratio with c and a.