Lesson 3, Topic 2
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# Angles in a Semicircle

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Theorem: An angle in a semicircle is a right angle.

Given: AB is a diameter of a circle with centre O, P is any other point on the circle.

To prove: âˆ APB = 90Â°

### Proof:

Join the radius PO, and let x = âˆ A and y = âˆ B.

Î”AOP and Î”BOP are isosceles because all radii are equal. i.e OA = OP = OB = OP

âˆ´ âˆ APO = x (base angles of isosceles triangle)

âˆ BPO = y (base angles of isosceles triangle)

âˆ APO + âˆ BPO = âˆ APB

âˆ´ x + y = âˆ APB

Also âˆ OAP + âˆ OBP +( âˆ APO + âˆ BPO) = 180Â° (sum of angles in Î”APB)

âˆ´ x + y + (x + y) = 180Â°

2x + 2y = 180Â°

x + y = $$\frac{180}{2}$$

x + y = 90Â°

Remember x + y = âˆ APB

âˆ´ âˆ APB = 90Â°

### Example 1

A, B, C and D are points on the circumference of the circle, centre O. âˆ DOB is a straight line and âˆ DAC = 58Â°
Find âˆ CDB

Solution:

â‡’ $$\scriptsize \angle DBC = \angle DAC$$ (angles in the same segment)

âˆ´ $$\scriptsize \angle DBC = 58^o$$

$$\scriptsize \angle DCB = 90^o$$ (angle in a semicircle is a right angle)

Consider Î”DCB

â‡’ $$\scriptsize \angle CDB \: + \: \angle DCB \: + \: \angle DBC = 180^o$$ (sum of angles in Î”DCB)

â‡’ $$\scriptsize \angle CDB \: + \: 90^o \: + \: 58^o = 180^o$$

â‡’ $$\scriptsize \angle CDB \: + \: 148^o = 180^o$$

â‡’ $$\scriptsize \angle CDB = 180^o \: – \: 148^o$$

â‡’ $$\scriptsize \angle CDB = 32^o$$

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