Let $O \in BD$ be the circumcenter of triangle $\triangle ABC$ and $M$ be the midpoint of $AB$ . Then $MO$ is the perpendicular bisector of $[AB]$. Let $P=MO \cap AC$ . $O$ is circumcenter of $\triangle ABC$ then $\triangle OAC$ is isosceles and $\angle PAO=\angle PCO$. Since $O,P$, belong to the perpendicular bisector of $AB$ , we get $\angle PBO=\angle PAO$ . So, $\angle PCO=\angle PBO$ which means that $POBC$ is cyclic so $PO$ and $BC$ are antiparallel . But $BC \parallel AD$ (bases of trapezoid) and so, $PO$ and $AD$ are also antiparallel. Thus $POAD$ is also cyclic, and $\angle PDO=\angle PAO=\angle PBO$. Therefore triangle $\triangle PBD$ is isosceles and $P$ belongs to the perpendicular bisector of $[BD]$ . But $P$ also belongs to the perpendicular bisector of $[AB]$ which means that it is the circumcenter of triangle $\triangle ABD$ and the conclusion follows.

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Let $M,X,Y$ be the midpoints of $AB,AC,BD$. They are collinear because $BC \parallel AD$. Let $O$ be the circumcenter of $\triangle ABC$ and let the perpendicular to $BD$ at $Y$ meet $AC$ at $P$. Then $POXY$ is cyclic (diameter $PO$). So $$\measuredangle POX = \measuredangle PYX = \measuredangle PYO + \measuredangle OYX = \measuredangle PYO + \measuredangle BDA = \measuredangle PYO + \measuredangle CBO = \measuredangle CAB = \measuredangle MOX.$$Thus, $P$ is the intersection of the perpendicular bisectors of $AB,BD$, so it is the circumcenter of $\triangle ABD$. Consequently, $P$ lies on $AC$.

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very easy solution. since BD is a radius through B in triange ABC, angle CBD = 90 minus angle CAB But in the trapezium AD is parallel to BD and so angle CBD = angle ADB So in triangle ADB, angle CAB = 90 minus angle ADB. This proves that AC is a radius of triangle ABD passing through A and so circumcenter lies on diagonal AC. Proved