P20 from A Beautiful Journey through Olympiad Geometry 
Let $ABCD$ be a quadrilateral such that $∠BCA + ∠CAD = 180°$and $AB = AD + BC$. Prove that $∠BAC + ∠ACD = ∠CDA$
Source: Serbia 2014, Opstinski IA

I was able to solve this using Trigonometry and Solution of Triangle but I am unable to find a synthetic solution. Any hints or suggestions are welcome

 A: Hint: Find a geometric transformation that simplifies the conditions.
Naively / with wishful thinking:

*

*Something becomes a straight line

*Two line segments have equal length.

*Two angles are placed side by side.

*Two angles are equal.

*2 and 4 suggest that there's an isosceles triangle.



 Reflect $D$ about the perpendicular bisector of $AC$ to point $D'$.

Show that the conditions imply that $BD'A$ is an isosceles triangle.
Hence, the statement follows.
$$\angle BAC + \angle ACD = \angle BAC + \angle CAD' = \angle BAD' = \angle BD'A = \angle CDA.$$
A: 
In figure we have:
$$DA=AE\Rightarrow \overset{\large\frown}{DA}=\overset{\large\frown}{AE}$$
$$\Rightarrow \widehat{DCA}=\frac{\overset{\large\frown}{AD}}{2}=\frac{\overset{\large\frown}{EC}}{2}$$
CF is rotated position of CB such that:
$$CF\bot OC\Rightarrow \angle ACF=\angle ACB=\frac{\overset{\huge\frown}{AEC}}2$$
$$\angle CAD=\frac{\overset{\large\frown}{AEC}}2=\frac{\overset{\large\frown}{AE}}2+\frac{\overset{\large\frown}{EC}}2=\angle DCA+\angle CAB$$
A: 
It is given that ${\angle BCA}+{\angle CAD}=180°$.
Consider $\triangle ABC$. ${\angle CAB}+{\angle ABC}={\angle CAD}$. $(1)$
It is given that AB=AD+BC.
Let 'M' be on AB such that AM=AD and MB=BC. So, ${\triangle AMD}$ and ${\triangle MBC}$ are isosceles.
Therefore, ${\angle BAD}+{\angle ABC}=360°-2({\angle ADM}+{\angle  BCM})=2{\angle CMD}$
And, form (1), ${\angle BAD}+{\angle ABC}=2{\angle DAC}$
So, ${\angle CMD}={\angle DAC}$
So, AMCD is cyclic.
So, ${\angle BAC}={\angle MAC}={\angle MDC}$, and,
${\angle ACD}={\angle AMD}={\angle ADM}$
Hence, ${\angle BAC}+{\angle ACD}={\angle CDM}+{\angle MDA}={\angle ADC}$
