If $CD$ is the diameter of the circumcircle of triangle $ABC$, then $A_1 = A$ and $B_1 = B$ and the statement holds. Assume that $CD$ is not the diameter (Fig. 3). Then $A_1 \ne A$ and $B_1 \ne B$. The point $A_1$ lies on the ray $AD$ if and only if the point $B_1$ does not lie on the ray $BD$ (depending on which side of the diameter through point $C$ point $D$ is located). Thus, $\angle CAA_1 =$
$\angle CBB_1$ (because the sum of opposite angles of a cyclic quadrilateral $ADBC$ is $180^\circ$). Thus, the right-angled triangles $AA_1C$ and $BB_1C$ are similar. From $\angle ACA_1 = \angle BCB_1$ we see that $\angle ACB = \angle A_1CB_1$. This together with $\frac{|AC|}{|BC|} = \frac{|A_1C|}{|B_1C|}$ gives that $ACB$ and $A_1CB_1$ are similar. As $|AC| > |A_1C|$, we conclude that $|AB| > |A_1B_1|$.

Solution 2:

Since the angles $CA_1D$ and $CB_1D$ are right angles, the points $C, A_1, D$, and $B_1$ form a cyclic quadrilateral and thus $\angle CAB = \angle CDB = \angle CA_1B_1$. Similarly, $\angle CBA = \angle CB_1A_1$. Therefore the triangles $ABC$ and $A_1B_1C$ are similar. As $|CA| \ge |CA_1|$, we deduce that $|AB| \ge |A_1B_1|$.

Fig. 3

Solution 3:

The radius $R$ of the circumcircle of the quadrilateral $CADB$ is at least as large as the radius $R_1$ of the circumcircle of the quadrilateral $CA_1DB_1$ because $CD$ is a chord in the first one and a diameter in the second one. The sine law in triangles $ADB$ and $A_1DB_1$ gives $|AB| = 2R \sin \angle D$ and $|A_1B_1| = 2R_1 \sin \angle D$. As $R \ge R_1$, we deduce $|AB| \ge |A_1B_1|$.