Indija TS

Let the circum-circle of $BCD$ be $\Gamma$ and let the circle with diameter $OD$ be ${\Gamma }^{\prime }$. Let the mid-point of $OD$ be $O^{\prime }$ (which is also the centre of ${\Gamma }^{\prime }$). Now $O{O}^{\prime }$ passes through $D$ and hence ${\Gamma }^{\prime }$ and $\Gamma$ are tangent to each other at $D$. Hence there is a homothety with centre $D$ taking ${\Gamma }^{\prime }$ to $\Gamma$, with ratio $2$. Since $\angle OMD=90^{\circ }$, the point $M$ is on ${\Gamma }^{\prime }$. Let the image of $M$ under the homothety be $D^{\prime }$; it lies on $\Gamma$ and $M$ is the midpoint of $D{D}^{\prime }$.

Consider the half-turn centred at $M$ taking $D$ to $D^{\prime }$. This also takes $C$ to $E$ as $M$ is the mid-point of $CE$. Let $B^{\prime }$ be the image of $B$ under this transformation. Since the half turn takes a line to another line parallel to it, the image of $BC$ is $B^{\prime }E$; and $BC\parallel B^{\prime }E$. Thus $AE\parallel BC\parallel B^{\prime }E$. It follows that $A$, $B^{\prime }$, $E$ are collinear. Observe that $A$, $B$ lie on the same side of $CE$. Since we have performed a rigid transformation preserving $C$, $E$, the images of $A$, $B$ must also lie on the same side of $CE$. This implies that $A$ and $B^{\prime }$ lie on different sides of $CE$. Thus $E$ lies between $A$ and $B^{\prime }$. Because of this, we have $$\begin{array}{rl}A{B}^{\prime }=AE+E{B}^{\prime }& =AE+BC(\text{half-turn takes }BC\text{ to }{B}^{\prime }E)\\ & =AB.\end{array}$$ We also observe that the half-turn takes triangle $C{D}^{\prime }B$ to $ED{B}^{\prime }$. Therefore $\angle C{D}^{\prime }B=\angle ED{B}^{\prime }$. Thus we get $$\begin{array}{rl}\angle BD{B}^{\prime }=\angle BDE+\angle ED{B}^{\prime }& =\angle BDE+\angle C{D}^{\prime }B=\angle BDE+\angle CDB\\ & =\angle CDE=\angle CBA=180^{\circ }-\angle BA{B}^{\prime },\end{array}$$ since $BC\parallel AE$. This shows that $B$, $D$, $B^{\prime }$, $A$ are concyclic. Since $AB=A{B}^{\prime }$, they subtend equal angle at $D$. Thus $\angle AD{B}^{\prime }=\angle BD{A}^{\prime }$. Thus $$2\angle BDA=\angle BDA+\angle AD{B}^{\prime }=\angle BDA+\angle ADE+\angle ED{B}^{\prime }.$$ But observe $\angle ED{B}^{\prime }=\angle C{D}^{\prime }B=\angle CDB$. Using this we obtain $$2\angle BDA=\angle BDA+\angle ADE+\angle CDB=\angle CDE.$$ This completes the proof.