prove that an angle in a semicircle is a right angle
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Given: AB is a diameter of a
circle with centre C
D is any point on the
circumference
To Prove: m∠ADB = 90°
Construction: Join CD
Proof: ACD is isosceles (AC = DC: radii)
⇒ m∠DAC = m∠ADC = α (base ∠s of an isosceles are equal) :
BCD is isosceles (BC = DC: radii)
⇒ m∠DBC = m∠BDC = β (base ∠s of an isosceles are equal)
However, the sum of the angles of ADB is 180°
⇒ α + α + β + β = 180°
⇒ α + β = 90°
⇒ m∠ADB = 90°
circle with centre C
D is any point on the
circumference
To Prove: m∠ADB = 90°
Construction: Join CD
Proof: ACD is isosceles (AC = DC: radii)
⇒ m∠DAC = m∠ADC = α (base ∠s of an isosceles are equal) :
BCD is isosceles (BC = DC: radii)
⇒ m∠DBC = m∠BDC = β (base ∠s of an isosceles are equal)
However, the sum of the angles of ADB is 180°
⇒ α + α + β + β = 180°
⇒ α + β = 90°
⇒ m∠ADB = 90°
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