jmc

Looking at triangle $ABD$, we see that $\angle BAD=180^{\circ }-\angle ABD-\angle ADB=180^{\circ }-40^{\circ }-55^{\circ }=85^{\circ }$. Then $\angle ABD<\angle ADB<\angle BAD$, so $AD<AB<BD$.

Looking at triangle $BCD$, we see that $\angle BCD=180^{\circ }-\angle CBD-\angle BDC=180^{\circ }-75^{\circ }-55^{\circ }=50^{\circ }$. Then $\angle BCD<\angle BDC<\angle CBD$, so $BD<BC<CD$.

Combining both inequalities, we see that $$AD<AB<BD<BC<CD.$$Therefore, the longest segment is $\boxed{CD}$.