Firstly we will show that the triangle is acute. Indeed, suppose that $\angle BAC \ge 90^\circ$. Then $\angle ACB \le 45^\circ = \angle ABC < \angle AA_1C$. From $\triangle AA_1C$ we get that $AC > AA_1$. From the other side

$\angle C_1AC \ge 90^\circ$, then in $\triangle AC_1C$ the side $CC_1$ is the longest, in particular, $CC_1 > AC$. Hence $CC_1 > AC$; $CC_1 > AC > AA_1$, contradiction.
Let $AA_1$ and $CC_1$ intersect at $O$. Consider a circle with diameter $AC$, then $O$ lies on it (fig. 10).

Fig. 10

This circle intersect $AB, BC$ at $P, Q$ respectively. The points $P$ and $Q$ lies inside the sides of $\triangle ABC$ because the triangle is acute. Notice also that since $AA_1$ and $CC_1$ are cevians, $O$ lies inside $\triangle ABC$, in other words, on the arc $PQ$ of the latter circle. We will prove that $\angle A_1AQ = \angle QAC$. If it is true $\triangle CAA_1$ is isosceles, because $AQ$ is angle bisector and altitude in it and so $A_1A = AC$. Indeed, suppose that $\angle A_1AQ = \beta > \alpha = \angle QAC$. Then $A_1A > AC$. Obviously, $\angle PBC = \angle BAQ = 45^\circ$. Then
$$\angle PCC_1 = 45^\circ - \angle OCQ = 45^\circ - \angle OAQ = 45^\circ - \beta.$$
Also $\angle PCA = 90^\circ - \angle PAC = 45^\circ - \alpha$. Since $\beta > \alpha$, we have $45^\circ - \beta < 45^\circ - \alpha$, so $\angle C_1CP < \angle PCA$, hence $CC_1 < AC$. But $A_1A > AC$, so $CC_1 < AA_1$ - contradiction. The case $\beta < \alpha$ is considered analogously.

Alternative solution. Denote $\angle C_1CB = \alpha$, $\angle A_1AB = \beta$. Then since $\angle OCA + \alpha + \angle OAC + \beta = 135^\circ$ and $\angle OCA + \angle OAC = 90^\circ$, so $\alpha + \beta = 45^\circ$ (fig. 11).

Fig. 11

We can assume that $AA_1 = CC_1 = 1$. Denote $OC_1 = p$, $OA_1 = q$, then $OC = 1-p$, $OA = 1-q$. Since $\tan \alpha = \frac{p}{1-p}$ and $\tan \beta = \frac{q}{1-q}$, rewrite the condition $\alpha + \beta = 45^\circ$ as $\tan(\alpha + \beta) = 1$ or
$$\begin{aligned} \tan(\alpha + \beta) &= \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\frac{p}{1-p} + \frac{q}{1-q}}{1 - \frac{p}{1-p} \frac{q}{1-q}} = \\ &= \frac{q - q^2 + p - p^2}{(1-p)(1-q) - pq} = \frac{q + p - (q^2 + p^2)}{1 - p - q} = 1, \end{aligned}$$
so $p^2 + q^2 - 2p - 2q + 1 = 0$. Now we have that
$$AC^2 = (1-p)^2 + (1-q)^2 = 2 - 2p - 2q + p^2 + q^2 = 1 = AA_1^2$$
and we are done.