What is the angle $\widehat{BAC}$? Let the triangle $ABC$ and the angle $\widehat{ BAC}<90^\circ$ 
Let the perpendicular to $AB$ passing by the point $C$ and the perpendicular to $AC$ passing by $B$ intersect the circumscribed circle  of $ABC$ on $D$ and $E$ respectively . 
We suppose that $DE=BC$
What is the angle $\widehat{BAC}$ 
I tried using law of sines in triangle
Also , let O be center of circle so OD=OE=r
 A: 
Note that $\angle BAC=\angle BEC$, and that $\angle BAC=180^\circ-\angle DFE=\angle CFE$. As $DE=BC$, $\angle BEC=\angle DCE$. Therefore, $\angle BAC = 60^\circ$.
Appendix:

See the comment section below.
A: Let $AECFBD $ (labeled anti-clockwise) be a regular hexagon inscribed in a circle center $O$. All the given conditions are satisfied by this assumption ( perpendicularities) and $(DE=BC).$  Simply by symmetry, angle BAC = $60^0.$ 
A: 
The above is basically what you're given. I've labeled a lot of the angles with $\alpha, \beta, \gamma, \delta$ and $\epsilon$. Two inscribed angles are equal when they're based off the same chord, ie I'm leaning heavily on the third diagram here.
Now note that $\alpha + \beta + \gamma + \delta + \epsilon = \pi$ and $\gamma + \delta = \gamma +\epsilon = \pi/2$. Furthermore, $\triangle BPC \cong \triangle DPE$ (angle side angle), so $\triangle BED \cong \triangle DCB$ (side side side) and we conclude $\gamma = \epsilon +\delta$. Therefore, $$\pi = (\gamma + \delta) + (\gamma + \epsilon) = 3\gamma$$ and so $\gamma = \pi/3$.
