Giving reasons , Find the following angles & show that $\triangle ODE$ is isosceles 
I don't see any angles in terms of $x$ ,  or why $\triangle ODE$ is isosceles I would highly appreciate your hints
 A: $\angle AOB = 2\angle AXB$ since $\angle AXB$ is the angle subtended at the circumference and $\angle AOB$ is subtended at the center by the chord $AB$. Also $\angle ABO = \frac{1}{2}(180^\circ - 2x)$, since $AOB$ is isosceles. Since $DB = DA$ and $OB=OA$, the triangles $DOB$ and $DOA$ are congruent. Thus $\angle DOB = x$. $\angle ABO = 2\angle BOE$ and hence $\angle BOE = \frac{1}{4}(180^\circ - 2x)$ and finally $BEO$ is right angled at $B$ and hence $\angle BEO = 90^\circ - \angle BOE$.
A: The triangle OAX is isosceles.
Likewise the triangle OAB is isosceles.
OA is perpendicular to DA.
OB is perpendicular to BD.
This should give you the angles DAB and DBA.
You should be able to at this point discern that length of AD equals length DB
A: Hint:
i): use the central angle theorem.
ii) the triangle $AOB$ is isosceles.
iii) $OD$ is orthogonal to $AB$.
iv) the angle $OBC$ is external to the triangle $AOB$ and the triangle $OBC$ is isosceles.
v) the triangle $OBE$ is rectangle in $B$.
Now it easy to prove vi).
