A relation between sides of quadrilateral

Question

Suppose $ABCD$ a quadrilateral (in Euclidean geometry), with $$ \overline{AD} \leq \overline{AC},\\ \overline{BC} \leq \overline{BD}. $$ Then show that $$ |\overline{AD} - \overline{BC}| \leq \overline{AB}. $$ (I will soon post a proof of mine which I am not sure whether it's completely rigorous. This is intended as a lemma for my proof of Existence of an 1-Lipschitz function.)

Answer 1

We need to prove that $$-AB<AD-BC<AB.$$ Indeed, $$AD+AB>BD>BC$$ and we got a left inequality and $$AB+BC>AC>AD$$ and we got a right inequality.

Done!

Answer 2

Now, draw a circle $C_A$ with center $A$, radius $\overline{AD}$, and draw a circle $C_B$ with center $B$, radius $\overline{BC}$.

Suppose $C_A,C_B$ intersect at one or two points, let any of the two points be $E$. Then $$ |\overline{AD}-\overline{BC}| =|\overline{AE}-\overline{BE}| \leq \overline{AB} $$ because of property of $\Delta ABE$.

Suppose $C_A,C_B$ do not intersect. Then $C_A$ and $\overline{AB}$ must intersect; call it $E$. Similarly let $C_A$ and $\overline{AB}$ intersect at $F$. So, $$ |\overline{AD}-\overline{BC}| =|\overline{AE}-\overline{BF}| \leq |\overline{AB}-0|. $$