Steps for the Proof:

1. Draw the Circle and the Triangle
2. Understand the Relationship
3. Consider the Angles
4. Consider the Angle Sum in Triangle $\ OAC$ and$\ OBC$

Draw the Circle and the Triangle:

• Draw the circle with centre $\ O$ and diameter $\ AB$.

• Let $\ C$ be any point on the circle, forming the triangle $\ ABC .$ Understand the Relationship:

• The points $\ A , \ B ,$ and $\ C$ are all on the circumference of the circle.

• $\ AB$ is the diameter, so the line$\ OC$ is the radius of the circle.

• Note that $\ OA = OB = OC = r$ (where $\ r$ is the radius of the circle). Consider the Angles:

• The angle $\ \angle OCA$ and$\ \angle OCB$ are both angles at the centre and subtend the same arc $\ AB .$

• Therefore, $\ \angle OCA = \angle OCB .$ Consider the Angle Sum in Triangle $\ OAC$ and$\ OBC :$

• In $\ \triangle OAC$ and $\ \triangle OBC :$ $\angle AOC + \angle ABC = 180^\circ \quad (\text{angles on a straight line})$ $\angle AOC = 2 \times \angle ACB$ But since $\ \angle AOC = 180^\circ$, it follows that: $2 \times \angle ACB = 180^\circ$ $\angle ACB = :highlight[90^\circ]$ This shows that $\ \angle ACB$ is indeed a right angle.

Given a circle with a diameter $\ AB = :highlight[10]$ units and a point $\ C$ on the circle such that $\ AC = :highlight[6]$ units and $\ BC = :highlight[8]$ units, verify that the angle $\ \angle ACB$ is a right angle.

Solution: 5. Check the Condition for a Right Triangle:

• To verify $\ \angle ACB = :highlight[90^\circ] ,$ check whether $\ \triangle ABC$ satisfies the Pythagorean theorem. $AC^2 + BC^2 = AB^2$
• Calculate: $6^2 + 8^2 = 36 + 64 = 100 = 10^2$
• Since $\ AC^2 + BC^2 = AB^2 , \ \triangle ABC$ is a right triangle.

6. Conclusion:

• By Thales' theorem, $\ \angle ACB = :highlight[90^\circ] ,$ confirming the result.

Given a circle with a centre at $O$ and diameter $\ PQ$, if $\ R$ is a point on the circumference such that $\ PRQ$ forms a triangle, what is the measure of $\ \angle PRQ$?