Let $AC = AB = R$ and $CC_1 = x$ and $BB_1 = y$ and $AA_1 = z$. Claim $:$ one of $x,y,z$ is sum of the other two. Proof $:$ Note that $x = \sin{COY} . R$ and $y = \sin{60 - COY} . R$ and $z = \sin{60+COY} . R$ WLOG assume $z = x+y$. Let $S$ be point on $AA_1$ such that $AS = BB_1 = y$. we will prove $B_1X = A_1O + C_1Y$ or $OB_1 = C_1A_1$ or $CS = OB_1$. Note that $AC = R = BO$ and $AS = BB_1 = y$ and $\angle ASC = \angle AA_1C_1 = \angle 90 = \angle BB_1O$ so $ASC$ and $BB_1O$ are congruent so $CS = OB_1$ as wanted so we're Done.

Since $ABOC$ is a rhombus, the projections of $AC$ and $BO$ onto $XY$ are equal, that is $A_1C_1=B_1O \Rightarrow B_1A_1=OC_1$. Therefore, $XO=OY\Rightarrow XB_1+B_1A_1+A_1O=OC_1+C_1Y \Rightarrow XB_1+A_1O=C_1Y$.