A quadrilateral and circularity of vertices So we have a convex quadrilateral ABCD, which satisfies these conditions: 
$m(\widehat{DAC})=m(\widehat{BDC})=36°$ and $m(\widehat{BAC})=72°$.
If $P=AC\cap BD, \,m(\widehat{APD})=\,?$
I did a quick sketch:

Observing the angles we can see that $2m(\widehat{BDC})=m(\widehat{BAC})$ and $2m(\widehat{DBC})=m(\widehat{CAD})$ which perfectly satisfy the conditions of a circle, whose center is $A$ and radii are $AB, \,AC$ and $AD$. But it also requires them to be equal.
My question is, do they have to be equal? If not, how can I solve this problem?
 A: Your insights about circles are very good. Here's a way to think about the diagram ...

Begin with segment $\overline{CD}$. Point $B$ lies somewhere on an arc of a circle  with center, say, $K$ such that $m\angle CKD = 36^\circ$. (There are two such circles. Pick one.) I'll call the arc of the circle on which $B$ lies the "inscribing arc for $18^\circ$". Clearly, there is only one position for $B$ such that $m\angle BDC = 36^\circ$.
Point $A$ is on the "inscribing arc for $36^\circ$" on segment $\overline{CD}$, but it's also on the "inscribing arc for $72^\circ$" on segment $\overline{BC}$; its position is uniquely determined as the (non-$C$) point of intersection of the two arcs. (Note that there's no choice in the arcs here, as their centers must lie on appropriate sides of segments $\overline{CD}$ and $\overline{BC}$.) 
Thus, given $\overline{CD}$, the configuration is (apart from the choice of that first circle) unique. As you have noted, the configuration in which $A$ is the center of a circle containing $B$, $C$, $D$ fits the given information; consequently, your configuration is the configuration, so you may safely assume that $\overline{AB}\cong\overline{AC}\cong\overline{AD}$ to solve the problem.
A: let $O$ be the circumcenter of $\triangle BCD$ and then $\angle BOC=72^{\circ}$, $\angle COD=36^{\circ}$
thus, $\square BCOA$ and $\square CDAO$ are cyclic, where their circumcircles are $(O_{1})$,$(O_{2})$ respectively.
that is, two circles meet at three points $A$,$C$,$O$, which is contradiction.
hence, $A$ must be identical with $O$.
