Find The Angle

If I were in school, my math teacher won't be happy with my style of answer:

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Let the point where $BO$ and $AC$ meet be $F$ .

Since $M$ is the midpoint of $DE$ , by properties of similar triangles, it can be proven that $F$ is the midpoint of $AC$ .... $\rightarrow Fact \space 1$

Further, since $A, M, \And \space C$ are points on a circle (with center $O$ ), and, by the properties of chords in a circle, and also considering $Fact \space 1$ , we get, $OF \perp AC$ ... $\rightarrow Fact \space 2$

Using $Facts 1 \space and \space 2$ , we can clearly understand that $\triangle AMC$ is isosceles, i.e. $AM = MC$ , and $MF \perp AC$ .

It clearly follows for any point $M'$ on $\vec{FM}$ (as long as its not at infinity), $AM' = CM'$ .

$B$ is also one such point on $\vec{FM}$ ; hence $AB = BC$ .

Therefore right $\triangle ABC$ is also isosceles. It is also right-angled at $B$ .

Hence $\angle BAC = 45^{\circ}$

Okay, so this is cheating, but the diagram has to be symmetrical through the line OB, because there are no distinguishing features in the question between points A and C and points D and E.

If it is symmetrical, ABC is an isoceles triangle, so angle BAC is 45 degrees.

I am aware that this is not how you are supposed to it. I call it initiative.