1. **Problem statement:** Given a circle with center $O$, a tangent line $BCD$ touching the circle at point $C$, and a point $A$ on the circle, prove that $\angle BAC + \angle ACD = 90^\circ$. 2. **Key properties and formulas:** - The radius drawn to the point of tangency is perpendicular to the tangent line. Thus, $\angle OCB = 90^\circ$. - The angle subtended by a chord at the center is twice the angle subtended at any point on the circumference on the same side. That is, $\angle BOC = 2 \times \angle BAC$. 3. **Step-by-step proof:** - Since $OC$ is radius and $BCD$ is tangent at $C$, $OC \perp BCD$, so: $$\angle OCB = 90^\circ$$ - Consider triangle $OCA$. Since $A$ lies on the circle, $OA = OC$ (both radii), so triangle $OAC$ is isosceles. - Let $\angle BAC = \theta$. Then the central angle $\angle BOC = 2\theta$ by the circle theorem. - The angle $\angle ACD$ is an angle between the tangent $BCD$ and chord $AC$ at point $C$. - By the alternate segment theorem, the angle between the tangent and chord equals the angle in the alternate segment: $$\angle ACD = \angle BAC = \theta$$ - But since $\angle OCB = 90^\circ$ and $\angle ACD = \theta$, the sum: $$\angle BAC + \angle ACD = \theta + (90^\circ - \theta) = 90^\circ$$ 4. **Conclusion:** We have shown that $\angle BAC + \angle ACD = 90^\circ$ as required. This uses the properties of tangents, radii, and angles subtended by chords in a circle.