We have $\angle{BCD}=\angle{BDC_1}$ and $\angle{BAD}=\angle{BDA_1}$. Then $\begin{aligned} \angle{ABC}+\angle{CA_1D}&=\angle{ABC}+{\angle{BDC_1}+\angle{BDA_1}\\ &=\angle{ABC}+\angle{BCD}+\angle{BAD}\\ &=\angle{ABC}+\angle{BCA}+\angle{CAB}=180^\circ \end{aligned}}$ which implies that $BA_1DC$ is cyclic. Thus, $\angle{BCD}=\angle{BDC_1}=\angle{BA_1C}$, which gives $A_1C_1\parallel{AC}$.

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We have $(DC_1,DA_1) \equiv (DC_1,DB)+(DB,DA_1) \equiv (CD,CB) + (AB,AD) \equiv (AB,BC) (mod \pi)$ That means $D, C_1, B_1, B$ are concylic. So we obtain: $(A_1B,A_1C_1) \equiv (DB,DC_1)\equiv(BC,CA) (mod \pi)$ $\Rightarrow A_1C_1 \| AC$

My general problem and solution dedicate to phuongtheong Problem. Let $ABC$ be a triangle and $D$ is a point. Circumcircle of triangle $DCA,DAB$ cut $BC$ again at $E,F$, respectively. $DE,DF$ cut $CA,AB$ at $P,Q$, respectively. Prove that $PQ\parallel BC$. Proof. From cyclic quadrilateral $ADEC$ and $ADFB$ we have \[(AP,AQ)=(AP,AD)+(AD,AQ)=(AC,AD)+(AD,AB)=(EC,ED)+(FD,FB)=(BC,DQ)+(DP,BC)=(DP,DQ)(\bmod\pi).\] This means $QPDQ$ is cyclic. Therefore \[(PQ,BC)=(PQ,AC)+(AC,BC)=(PQ,PA)+(CA,CE)=(DQ,DA)+(DA,DE)=0(\bmod\pi).\] So $PQ\parallel BC$, we are done.

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Draw in line segment $BD$. Note that $\angle BDA_1=\angle BAD$ and $\angle BDC_1=\angle BDC$, so $\angle CDA_1=\angle BAC+\angle BCA=180^\circ-\angle ABC$. Therefore $BA_1DC_1$ is cyclic, so $\angle BCA=\angle C_1DB=\angle C_1A_1B$ and $A_1C_1\parallel AC$ as desired. $\blacksquare$

$\angle C_1DB=\angle C$,so $\angle C_1DA=\angle CBD$,similarly get $\angle CDA_1=\angle ABD$,but $\angle ADC=180$,so $BA_1DC_1$ is cyclic.but then $\angle BC_1A_1=\angle A$,hence done.

Dear Mathlinkers, 1. the angled line between the two tangents at D and <CBA are supplementary; in consequence, B, A1, D and C1 are concyclic 2. This circle going through B and D is a circle of constant ratio 3. by Thales theorem, A1C1 // AC. Sincerely Jean-Louis

Sorry this is probably a dumb question but why is $\angle BDA_1=\angle BAD$ and $\angle BDC_1=\angle BDC$

We use $A$-centered labelling because $B$-centered labelling ew. The main claim is that $AB_2DC_1$ is cyclic, which is obvious as $\angle ADB_2 = B$ and $\angle ADC_2 = C$. Then $$\angle AC_2B_2 = \angle ADB_2 = \angle B,$$so $\overline{B_1C_1} \parallel \overline{BC}$, as required.

since we have $\angle{C_{1}DB}=\angle{BCD}$ and $\angle{A_{1}DB}=\angle{BAD}$ so we have $\angle{C_{1}DA_{1}}=180-\angle{C_{1}BA_{1}}$ hence we have points $B,C_{1},D,A_{1}$ to be concyclic hence we have $\angle{BC_{1}A_{1}}=\angle{BDA_{1}}=\angle{BAC}$ and $\angle{BA_{1}C_{1}}=\angle{BDC_{1}}=\angle{BCA}$ hence we have $A_{1}C_{1}|| AC$ $\blacksquare$

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thugzmath10 wrote: We have $\angle{BCD}=\angle{BDC_1}$ and $\angle{BAD}=\angle{BDA_1}$. Then \begin{align*} \angle{ABC}+\angle{CA_1D}&=\angle{ABC}+{\angle{BDC_1}+\angle{BDA_1}}\\ &=\angle{ABC}+\angle{BCD}+\angle{BAD}\\ &=\angle{ABC}+\angle{BCA}+\angle{CAB}=180^\circ \end{align*}which implies that $BA_1DC$ is cyclic. Thus, $\angle{BCD}=\angle{BDC_1}=\angle{BA_1C}$, which gives $A_1C_1\parallel{AC}$. ftfy