A circle is circum-scribed of ΔABC. O is the center of the circle and CP is the tangent of the circle at the point C. If BC bisects the ∠OCP, then what is the supplementary angle of ∠BAC?

Step 1 (Tangent and Radius Angle): Since $CP$ is a tangent to the circumcircle at $C$ and $OC$ is the radius to the point of contact, we have:

Step 2 (Angle Bisector): We are given that $BC$ bisects $\angle OCP$. Thus:

Step 3 (Find Central Angle BOC): In the triangle $△OBC$, $OB$ and $OC$ are both radii of the circle, so $OB=OC$. This means $△OBC$ is an isosceles triangle, so:

Using the angle sum of $△OBC$:

Step 4 (Find Inscribed Angle BAC): By the Inscribed Angle Theorem, the angle $\angle BAC$ subtended at the circumference is half of the central angle $\angle BOC$ subtending the same arc $BC$:

Step 5 (Compute Supplementary Angle): The supplementary angle of $\angle BAC$ is:

Step 6 (Conclusion): The supplementary angle of $\angle BAC$ is exactly 135 degrees.