$A,B,C,D$ on a circle. $\widehat{BAC}=\widehat{BDC}$ Points $A,B,C,D$ belong to a circle. What's a rigorous yet simple proof that $\widehat{BAC}=\widehat{BDC}$ ?
Does this property have a name?


I get that in the above figure:


*

*summing angles, $\widehat{BOA}+\widehat{AOC}=\widehat{BOC}=\widehat{BOD}+\widehat{DOC}$

*sum of the angles of isosceles triangle $AOB$ is $\Pi$, thus$\widehat{BOA}=\Pi-2\widehat{BAO}\quad$ and similarly
$\widehat{AOC}=\Pi-2\widehat{OAC}\quad$
$\widehat{BOD}=\Pi-2\widehat{BDO}\quad$
$\widehat{DOC}=\Pi-2\widehat{ODC}\quad$

*replacing then simplifying, we get
$\widehat{BAO}+\widehat{OAC}=\widehat{BDO}+\widehat{ODC}\quad$ thus
$\widehat{BAC}=\widehat{BDC}\quad$ Q.E.D.
However this reasoning seems dependent on the order of points on the circle, and perhaps other hypothesis.
 A: This is a well known property of angles, inscribed in a chord of a circle: they are congruent if they are inscribed on the same side of the chord (BC in your case) and supplementary if inscribed on opposite sides of the chord.
A: 
Here is a proof that an angle inscribed in an arc is half the intersepted arc.
That is, $m\angle APB=\frac{1}{2}m(arc \widehat{ACB})$
Following the figure, let $\mathrm{M}$ be the center of the circle, and 
$\angle APB=\alpha , \angle APM=\beta\\
\mathrm{Now, \: in \: \Delta MPB,}\\
\mathrm{MP}=\mathrm{MB}\\
\implies \angle MPB= \angle MBP= \alpha + \beta\\
\implies \angle CMB= 2 ( \alpha +\beta)\\
\: \\
\mathrm{In \: \Delta MPA,}\\
\mathrm{MP} =\mathrm{MA} \\
\angle MPA= \angle MAP= \beta \\
\implies \angle CMA=2\beta\\
\mathrm{Since \:} \angle CMB= \angle CMA+ \angle AMB\\
2(\alpha+ \beta)= 2\beta + \angle AMB\\
\implies 2\alpha=\angle AMB\\
\: \\
\mathrm{And, \: since\:} \angle APB=\alpha, \mathrm{we \: have \: shown \:that} \\
\boxed{\angle APB=\frac{1}{2} \angle AMB} $
Now, your statement, that is, angles inscribed in the same arc are congruent, is a direct consequence of the previous theorem. 
For in your case, 
$\angle BAC= \frac{1}{2}\angle BOC\\
\angle BDC=\frac{1}{2}\angle BOC\\ 
\implies \angle BAC=\angle BDC$
