Solution 1. (Based on work by Oleg Golberg) We start with an important geometric observation.

Lemma. Triangle $ABC$ inscribed in circle $\omega$. Lines $BT$ and $CT$ are tangent to $\omega$. Let $M$ be the midpoint of side $BC$. Then $\angle BAT = \angle CAM$. (Line $AT$ is a symmedian of triangle.)

Proof. We consider the above configuration. (If $\angle BAC$ is obtuse, our proof can be modified slightly.) Let $D$ denote the second intersection (other than $A$) of line $AT$ and circle $\omega$. Because $BT$ is tangent to $\omega$ at $B$, $\angle TBD = \angle TAB$. Hence triangles $TBD$ and $TAB$ are similar, implying that $BD/AB = TB/TA$. Likewise, triangles $TCD$ and $TAC$ are similar and $CD/AC = TC/TA$. By equal tangents, $TB = TC$. Consequently, we have $BD/AB = TB/TA = TC/TA = CD/AC$, implying that
$$BD \cdot AC = CD \cdot AB.$$
By the Ptolemy's theorem to cyclic quadrilateral $ABDC$, we have $BD \cdot AC + AB \cdot CD = AD \cdot BC$.
Combining the last two equations, we obtain that $2 BD \cdot AC = AD \cdot BC$ or
$$\frac{AC}{AD} = \frac{BC}{2BD} = \frac{MC}{BD}.$$
Further considering that $\angle ACM = \angle ACB = \angle ADB$ (since $ABDC$ is cyclic), we conclude that
triangle $ABD$ is similar to triangle $AMC$, implying that $\angle BAT = \angle BAD = \angle CAM$. $\square$

Because $BT$ is tangent to $\omega$, $\angle CBT = \angle CAB$, and so
$$\angle TBA = \angle ABC + \angle CBT = \angle ABC + \angle CAB = 180^\circ - \angle BCA.$$
By the lemma, we have $\angle BAT = \angle CAM$. Applying the Law of Sines to triangles $BAT$ and $CAM$, we obtain
$$\frac{BT}{AT} = \frac{\sin \angle BAT}{\sin \angle TBA} = \frac{\sin \angle CAM}{\sin \angle BCA} = \frac{MC}{AM}.$$
Note that $TB = TC_1$. Thus, $TC_1/TA = MC/MA$. By equal tangents, $TB = TC$. In isosceles triangle $BTC$, $M$ is the midpoint of base $BC$. Consequently, $\angle TMS = \angle TAC = \angle TAS = 90^\circ$, implying that $TMAP$ is cyclic. Hence $\angle AMC = \angle ATC_1$. Because
$$\frac{AM}{AT} = \frac{MC}{TC_1} \quad (*)$$
and $\angle AMC = \angle ATC_1$, triangles $MAC$ and $TAC_1$ are similar. Because $BC/BM = B_1C_1/TC_1 = 2$, triangles $ABC$ and $AB_1C_1$ are similar.

Solution 2. (By Alex Zhai) We maintain the notations in the first proof. As shown at the end of the first proof, it suffices to show that $(*)$.

Let $O$ be the circumcenter of $ABC$. Note that triangles $OMC$, $OCT$ are similar to each other, implying that $OM/OC = OC/OT$ or $OM \cdot OT = OC^2 = OA^2$. Thus triangles $OAM$ and $OTA$ are also similar to each other. Further note that triangles $OMC$ and $CMT$ are also similar to each other. These similarities (which amount to the circumcircle of $ABC$ being a circle of Apollonius) give
$$\frac{AM}{AT} = \frac{OM}{OA} = \frac{OM}{OC} = \frac{MC}{CT} = \frac{MC}{TB} = \frac{MC}{TC_1},$$
which is $(*)$.

Solution 3. (Based on work by Sherry Gong) We maintain the notations of the previous solutions. Let $\omega$ intersect lines $AT$ and $AS$ again at $X$ and $Y$ (other than $A$), respectively. Let lines $YB$ and $CX$ meet at $B_2$, and let $YC$ and $BX$ meet at $C_2$. Applying Pascal's theorem to cyclic (degenerated) hexagon $BBYAXC$ shows that intersections of three pairs of lines $BB$ and $AX$, $BY$ and $XC$, and $YA$ and $CB$ are collinear; that is, $B_2, C_2, S$ are collinear. Likewise, applying Pascal's theorem to cyclic (degenerated) hexagon $CCYAXB$ shows that $B_2, C_2, T$ are collinear. We conclude that $B_2, C_2, S, T$ are collinear.

Since $ACXY$ is cyclic, $\angle YCX = \angle YAX = 180^\circ - \angle XAB = 90^\circ$. Thus $\angle B_2CC_2 = \angle XCC_2 = 180^\circ - \angle YCX = 90^\circ$. Likewise, $\angle C_2BB_2 = 90^\circ$. It follows that $BCC_2B_2$ is inscribed in a circle with $B_2C_2$ as its diameter. Thus the circumcenter of this circle is the intersection of lines $ST$ and the perpendicular of segment $BC$. This circumcenter must be $T$, and consequently, $B_2 = B_1$ and $C_2 = C_1$.

Because $AYBC$ and $B_1C_1CB$ are cyclic, by Miquel's theorem, $SACC_1$. (Indeed, $\angle ACB = 180^\circ - \angle B_1YS$ and $\angle BCC_1 = 180^\circ - \angle YB_1S$ lead to $\angle ACC_1 = 360^\circ - \angle ACB - \angle BCC_1 = 180^\circ - YSB_1$.) Also, by Miquel's theorem, $YAC_1B_1$ is cyclic. (Indeed, $\angle C_1AX = \angle C_1CS = \angle SB_1Y$.) By these cyclic quadrilaterals, it is not difficult to obtain $\angle ACS = \angle AC_1S$ (or $\angle ACB = \angle AC_1B_1$) and $\angle ABC = \angle AYC = \angle AYC_1 = \angle AB_1C_1$. Consequently, triangles $ABC$ and $AB_1C_1$ are similar to each other.