Find the value of $\angle BDE$

Question

In the figure below, $\angle CAB=\angle ABC=\angle ADB=70^\circ$ and $CD=BE$. Find $\angle BDE$.

Answer 1

Let $O$ be the center of the circumcircle $\Omega$ of the triangle $BCD$. Then, $OC=OD=OB$. Because $$\angle BCD=40^\circ=\angle ABD\,,$$ we conclude that $AB$ is tangent to $\Omega$ at $B$. Therefore, $OB\perp AB$, so$$\angle CBO=90^\circ-\angle ABC=20^\circ\,.$$ Because $$\angle CBD=\angle ABC-\angle ABD=70^\circ-40^\circ=30^\circ$$ and $$\angle COD=2\,\angle CED=60^\circ\,,$$ the triangle $COD$ is equilateral (having $OC=OD$ and $\angle COD=60^\circ$). Thus, the radius $\Omega$ is $OC=OD=CD=BE$.

In the figure above, all thick line segments have the same length. As $OB=BE$, the triangle $BOE$ is isosceles and we get $$\angle BOE=\frac{180^\circ-\angle EBO}{2}=\frac{180^\circ-20^\circ}{2}=80^\circ\,.$$ Now, $$\angle DOB=2\,\angle DCB=80^\circ=\angle BOE\,.$$ Ergo, $D$, $E$, and $O$ are collinear. Thus, $$\begin{align}\angle BDE&=\angle BDO=\angle DBO\\&=\angle CBD+\angle CBO=30^\circ+20^\circ=50^\circ\,.\end{align}$$