The circumcentre of the triangle $ABC$ is $O$. Prove that $\angle OBC + \angle BAC = 90^{\circ}$.

(N/A) $ABC$ is a triangle and $O$ is the circumcentre.
Draw $OD \perp BC$. Join $OB$ and $OC$.
In right $\Delta OBD$ and right $\Delta OCD$,we have:
hyp. $OB = \text{hyp. } OC$ [Radii of the same circle]
$OD = OD$ [Common side]
$\therefore \Delta OBD \cong \Delta OCD$ [By $RHS$ congruence rule]
$\therefore \angle 1 = \angle 2$ and $\angle 3 = \angle 4$ [$CPCT$]
Now,$\angle BOC = 2 \angle BAC$ [Angle at the centre is double the angle at the circumference]
Also,in $\Delta OBC$,$OB = OC$,so $\angle 3 = \angle 4$.
In $\Delta OBD$,$\angle 3 + \angle 1 + 90^{\circ} = 180^{\circ} \Rightarrow \angle 3 + \angle 1 = 90^{\circ}$.
Since $\angle BOC = 2 \angle 1 + 2 \angle 2$ and $\angle BOC = 2 \angle BAC$,we have $\angle 1 + \angle 2 = \angle BAC$.
Since $\angle 1 = \angle 2$,we have $2 \angle 1 = \angle BAC$,or $\angle 1 = \frac{1}{2} \angle BAC$.
Substituting this into $\angle 3 + \angle 1 = 90^{\circ}$,we get $\angle OBC + \angle BAC = 90^{\circ}$ (where $\angle OBC = \angle 3$).
Hence,proved.