keep seeing the attachment whenever necessary: Let $E$ be a point on $BC$ such that $BE=BC$ and $AE=AD$ (given $AB=BC+AD$) Obviously join $EC$and $ED$ Construct the angle bisector of $\angle A$ this is perpendicular to $ED$($\triangle EAD$ is isosceles) Mark $\angle B$ as $2x$ thus, $\angle BEC$ and $\angle BCE =90-x$ now $\angle A=180-2x$ ($BC$ is parallel to $AD$) thus, $\angle AED=x$ thus, $\angle CED=90$ thus $EC$ is parallel to the bisector of $\angle A$ Apply mid-point theorem to $\triangle ECD$ to get "the bisector of $\angle A$ also bisect $CD$".

Attachments:

Let the bisector of $ \angle A$ cut $CD$ at $M$ and $BC$ at $Q$. As $AD\parallel BC$ we conclude that $\angle DAQ=\angle AQB$ and since $AQ$ is the bisector of the $ \angle A$ then $\angle AQB=\angle BAQ$ and so $CQ=AD$ then the $\triangle AMD=\triangle CMQ$ and so $M$ is the middle of $DC$

Let the bisector of $A$ meets $CD$ and $BC$ at $L,M$,clearly $BA=BM$. so $CM=AD$,hence $ALD$ and $CLM$ are congruent.hence done.

Let the angle bisector meet DC at M and BC produced at N.. Iff M is mid point of DC then triangle ADM is congruent to triangle NCMSo angle BAM=a/2=ang MNC But since angle BAN is also a/2, It is enough to prove that triangle ABN is isosceles. We are given that AB= AD +BC, a construction such that BC be produced to X such that BX=AB will prove the required.

My solution: Denote $AB,AD,BC$ by $a+b,a,b$ where $b>a$.Extend $AB$ and $CD$ to meet at $E$.Since,$\bigtriangleup AED$ is similar to $\bigtriangleup BEC$,with some algebra,we find that $AE$=$\frac{ab+a^2}{b-a}$.Let the angle bisector of $\angle A$ meet $CD$ at $F$.Notice that $AF$ is the external bisector of $\angle DAE$.Hence,by applying External Angle Bisector Theorem on $\bigtriangleup AED$,we get that $\frac{EF}{DF}$=$\frac{AE}{AD}$=$\frac{b+a}{b-a}$ $\rightarrow$ $\frac{ED}{DF}$=$\frac{2a}{b-a}$.Again,from the similarity of $\bigtriangleup AED$ and $\bigtriangleup BEC$,we find $\frac{ED}{CD}$=$\frac{a}{b-a}$.This calls for a party because we are almost done:d It implies $\frac{2.ED}{CD}$=$\frac{ED}{DF}$ $\rightarrow$ $\frac{CD}{DF}$=$2$ or $\frac{CF}{DF}$=$1$,hence $AF$ bisects $CD$. (Done with length bashing).