Prove that $\angle AED+\angle ADO=90^\circ$ given that $\angle BAC=60^\circ$. Let ABC be a triangle such that $\widehat{BAC}=60^\circ$ and $AB\neq AC \neq BC$ and $O$ is the circumcenter of $ABC$.
Let $D$ be the intersection of the internal angular bisector of $\widehat{BAC}$ and $(BC)$.
And $E$ the intersection of $(OA)$ and $(BC)$
prove that $\widehat{AED}+\widehat{ADO}=90°$
I find the value of $\widehat{AED}$ and $\widehat{OAD}$ so i need $\widehat{AOD}$, i am not sure wheither am going on the right way or not

 A: I shall refer to your picture in my answer.  Your picture assumes that $\angle ABC<\angle ACB$, but the proof is similar if $\angle ABC>\angle ACB$.  That is, $H$ is the second intersection of the line $AD$ with the circumcircle of the triangle $ABC$.
Observe that the triangle $OCH$ is equilateral (this part uses the assumption that $\angle BAC=\dfrac{\pi}{3}$).   As $OH\perp BC$, the line $BC$ is an axis of symmetry of the triangle $OCH$.  This proves that $OD=DH$.  Now, let $D'$ be the image of the reflection of $D$ about the line $OH$.  Then, $D'$ is on $BC$ and the quadrilateral $ODHD'$ is a rhombus.  Therefore, $OD\parallel D'H$.  This means $$\angle ODA=\angle DHD'\,.$$
Note that the triangle $BHD'$ and the triangle $CHD$ are congruent.  That is,
$$\angle BHD'=\angle CHD=\angle CHA=\angle ABC\,.$$
On the other hand,
$$\angle AHB=\angle ACB\,.$$
Therefore,
$$\angle DHD'=\angle AHB-\angle BHD'=\angle ACB-\angle ABC\,.$$
This proves that
$$\angle ADO=\angle ACB-\angle ABC\,.$$
It is easy to see that $$\angle AED=\frac{\pi}{2}-\angle ACB+\angle ABC\,.$$
Ergo,
$$\angle AED+\angle ADO=\frac{\pi}{2}\,,$$
as desired.
