Denote by $P$, $Q$, and $R$ the centroids of the triangles $ABD$, $BCD$, and $ADE$, respectively. Let $K$ and $L$ be the midpoints of the segments $BD$ and $AD$, respectively. Since $P$ and $Q$ are centroids, they divide in the same ratio the medians $AK$ and $CK$, respectively. That is, $AP : PK = CQ : QK = 2 : 1$, and we have $PQ \parallel AC$. Similarly $PR \parallel BE$. Hence the angle $QPR$ is of the same measure as the angle $CXE$ determined by the lines $AC$ and $BE$ (here, $X$ is the intersection point of $AC$ and $BE$, see Fig. 1).

Fig. 1

Denote by $\varphi$ the measure of the inscribed angle determined by the chord $AB$ of the given circle. Since $CD = DE = AB$, we have $\angle CAE = 2\varphi$, and therefore from the triangle $AXE$ we conclude
$$\angle CXE = 180^\circ - \angle AXE = \varphi + 2\varphi = 3\varphi.$$
Since $AB > r$, we have $\varphi > 30^\circ$, and so $\angle QPR = 3\varphi > 90^\circ$.