A quadrilateral $ABCD$ is such that the sides $AB$ and $DC$ are parallel, and $|BC| =|AB| + |CD|$. Prove that the angle bisectors of the angles $\angle ABC$ and $\angle BCD$ intersect at right angles on the side $AD$.
Let $X \in BC$ such that $AB=BX; CX=CD$
Let $\angle ABX=\alpha => \angle XCD=180-\alpha $
since $\Delta ABX, \Delta XCD$ are isosceles we get $ \angle AXD = 90; \angle BYC=90$ where $Y$ is where the angle bisectors intersect
For $Y \in AD$ we define $Y$ as $AD \cap l_b$ where $l_b$ is the angle bisector at $B$ and from there all we need is angle chasing ($YXCD$ kite)