1. **Problem statement:** Given a circle with center O and points A, B, C, D on the circumference, where OA is parallel to CD, and angles $\angle OAB = 23^\circ$ and $\angle OCB = 41^\circ$, find: (a) $\angle ABC$ (b) $\angle AOC$ (c) $\angle OAD$ 2. **Key facts and formulas:** - Angles subtended by the same chord in the same segment are equal. - The angle between a radius and a tangent is $90^\circ$ (not directly used here but useful for circle problems). - Parallel lines imply alternate interior angles are equal. - The sum of angles in a triangle is $180^\circ$. 3. **Find $\angle ABC$:** - Since OA is parallel to CD, and AB and BC are chords intersecting these lines, $\angle OAB$ and $\angle BCD$ are alternate interior angles. - Given $\angle OAB = 23^\circ$, so $\angle BCD = 23^\circ$. - $\angle OCB$ is given as $41^\circ$. - In triangle $BCD$, angles are $\angle BCD = 23^\circ$, $\angle OCB = 41^\circ$, and $\angle BDC$ unknown. - But $\angle ABC$ is the angle at B in triangle $ABC$. - Consider triangle $ABC$ with points on the circle. - The angle at B, $\angle ABC$, subtends arc $AC$. - The angle at O, $\angle AOC$, is the central angle subtending the same arc $AC$. - The inscribed angle theorem states $\angle ABC = \frac{1}{2} \angle AOC$. 4. **Find $\angle AOC$:** - To find $\angle AOC$, use the given angles and parallelism. - Since OA is parallel to CD, $\angle OAB = 23^\circ$ equals $\angle BCD = 23^\circ$. - $\angle OCB = 41^\circ$ is given. - In triangle $BCD$, sum of angles is $180^\circ$: $$\angle BCD + \angle BDC + \angle DBC = 180^\circ$$ - We know $\angle BCD = 23^\circ$, $\angle OCB = 41^\circ$ (which is $\angle DBC$), so: $$23^\circ + 41^\circ + \angle BDC = 180^\circ$$ $$\angle BDC = 180^\circ - 64^\circ = 116^\circ$$ - $\angle BDC$ subtends arc $BC$. - The central angle $\angle BOC$ subtending the same arc is twice $\angle BDC$: $$\angle BOC = 2 \times 116^\circ = 232^\circ$$ - Since $\angle AOC = \angle BOC - \angle AOB$, and $\angle AOB$ is unknown, we need another approach. 5. **Alternative approach to find $\angle AOC$:** - Use the fact that $\angle OAB = 23^\circ$ and $\angle OCB = 41^\circ$. - Triangles $OAB$ and $OCB$ are isosceles because OA = OB = OC = radius. - In triangle $OAB$, angles at A and B are equal, so: $$\angle OAB = \angle OBA = 23^\circ$$ - So angle at O in triangle $OAB$ is: $$180^\circ - 2 \times 23^\circ = 134^\circ$$ - Similarly, in triangle $OCB$, angles at C and B are equal: $$\angle OCB = \angle OBC = 41^\circ$$ - Angle at O in triangle $OCB$ is: $$180^\circ - 2 \times 41^\circ = 98^\circ$$ - $\angle AOC$ is the sum of angles at O in triangles $OAB$ and $OCB$ minus $\angle BOC$ (which is the angle between OA and OC). - Since $\angle AOB = 134^\circ$ and $\angle BOC = 98^\circ$, then: $$\angle AOC = 134^\circ + 98^\circ = 232^\circ$$ 6. **Find $\angle ABC$ using inscribed angle theorem:** $$\angle ABC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 232^\circ = 116^\circ$$ 7. **Find $\angle OAD$:** - Since OA is parallel to CD, $\angle OAD$ equals $\angle ADC$ (alternate interior angles). - $\angle ADC$ subtends arc $AC$. - $\angle ABC$ subtends the same arc $AC$ and is $116^\circ$. - Angles subtending the same arc are equal, so: $$\angle OAD = \angle ADC = \angle ABC = 116^\circ$$ **Final answers:** (a) $\angle ABC = 116^\circ$ (b) $\angle AOC = 232^\circ$ (c) $\angle OAD = 116^\circ$