Fit Triangle

Relevant wiki: Solving Triangles - Problem Solving - Medium

$\angle BDC = 180^\circ - \angle B - \angle BCD = 180^\circ - 78^\circ - 24^\circ = 78^\circ = \angle B$

Therefore, $\triangle BCD$ is isosceles and $BC=CD$ .

$\angle BEC = 180^\circ - \angle C - \angle CBE = 180^\circ - 78^\circ - 51^\circ = 51^\circ = \angle CBE$

Therefore, $\triangle BCE$ is isosceles and $BC=CE$ . This implies that $CD=CE$ and $\triangle CDE$ is isosceles with $\angle CDE = \angle CED = \dfrac{180^\circ-(\angle C - \angle BCE)}{2} = \dfrac{180^\circ-(78^\circ- 24^\circ)}{2} = 63^\circ$ ,

Therefore, $\angle BED = \angle CED - \angle BEC = 63^\circ - 51^\circ = \boxed{12}^\circ$

We have;
In Triangle $CEB$ , $\angle CEB = (180-78-51)^\circ$
i.e. $\angle CEB= 51 ^ \circ$
THEREFORE, $CE=CB$

Also, In triangle $DCB$ , $\angle CDB = (180-78-24)^\circ$
i.e $\angle CDB = 78 ^\circ$
THEREFORE, $CD=CB$

THEREFORE, $C$ is the circumcentre of triangle $BDE$
So, $\angle BED =$ $\frac{1}{2}\times \angle BCD$
THEREFORE, $\boxed{\angle BED = 12^\circ}$