AB and CD are line segments, $AB=CD=1$, intersecting in point O, $\enspace$ $AB\cap CD =O$, $\angle AOC=60^{\circ}$. Prove that $AC+BD\geq1$. $\enspace$
What I tried:
$AO+BO=1$, $\enspace$ $CO+DO=1$
$AO=x$, $\enspace$ $CO=y$
$BO=1-x$, $\enspace$ $DO=1-y$
Law of cosines: $AC^2=x^2+y^2-xy$ $\enspace$ (1)$\enspace$ and $\enspace$ $BD^2=x^2+y^2-xy-x-y+1$ $\enspace$ (2)$\enspace$
But expressing AC and BD through (1) and (2) and replacing them in $AC+BD\geq1$, I get nowhere.
Could someone give me an idea to prove it?
Let $ABDK$ be a parallelogram.
Thus, $AK=DB$, $AD=AB=DC=1$ and $\measuredangle ADC=\measuredangle AOC=60^{\circ}.$
Hence, $\Delta KDC$ is an equilateral triangle, which says $KC=1$ and by the triangle inequality $$AC+BD=KA+AC\geq KC=1.$$
Also, you can end your work by the triangle inequality again: $$AC+BD=\sqrt{x^2+y^2-xy}+\sqrt{(1-x)^2+(1-y)^2-(1-x)(1-y)}=$$ $$=\sqrt{\left(x-\frac{1}{2}y\right)^2+\frac{3}{4}y^2}+\sqrt{\left(1-x-\frac{1}{2}(1-y)\right)^2+\frac{3}{4}(1-y)^2}\geq$$ $$\geq\sqrt{\left(x-\frac{1}{2}y+1-x-\frac{1}{2}(1-y)\right)^2+\frac{3}{4}(y+1-y)^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=1.$$