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

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

To prove: ∠APB = 90°

Proof:

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

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ΔAOP and ΔBOP are isosceles because all radii are equal. i.e OA = OP = OB = OP

∠APO = x (base angles of isosceles triangle)

∠BPO = y (base angles of isosceles triangle)

∠APO + ∠BPO = ∠APB

x + y = ∠APB

Also ∠OAP + OBP +( ∠APO + ∠BPO) = 180° (sum of angles in ΔAPB)

∴ x + y + (x + y) = 180°

2x + 2y = 180°

x + y = \( \frac{180}{2}\)

x + y = 90°

Remember x + y = ∠APB

∠APB = 90°

Example 1

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

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Solution:

⇒ \( \scriptsize \angle DBC = \angle DAC \) (angles in the same segment)

∴ \( \scriptsize \angle DBC = 58^o \)

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

Consider ΔDCB

⇒ \( \scriptsize \angle CDB \: + \: \angle DCB \: + \: \angle DBC = 180^o \) (sum of angles in ΔDCB)

⇒ \( \scriptsize \angle CDB \: + \: 90^o \: + \: 58^o = 180^o \)

⇒ \( \scriptsize \angle CDB \: + \: 148^o = 180^o \)

⇒ \( \scriptsize \angle CDB = 180^o \: – \: 148^o \)

⇒ \( \scriptsize \angle CDB = 32^o \)

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