Given ABCD is a cyclic quadrilateral.
To prove that ∠A + ∠C = 180° and ∠B + ∠D = 180°
Construction: Join AC and BD.
Proof:
<em>Angles in the same segment are equal.</em>
∠ACD = ∠ABD – – – – (1)
∠DAC = ∠DBC – – – – (2)
Add (1) and (2), we get
∠ACD + ∠DAC = ∠ABD + ∠DBC
⇒ ∠ACD + ∠DAC = ∠ABC (Since ∠ABD + ∠DBC = ∠ABC)
Add ∠ADC on both sides.
⇒ ∠ACD + ∠DAC + ∠ADC = ∠ABC + ∠ADC
We know that <em>sum of the angles of a triangle is 180°.</em>
In ΔABC, ∠ACD + ∠DAC + ∠ADC = 180°
Now, 180° = ∠ABC + ∠ADC
180° = ∠B + ∠C – – – – (3)
<em>Sum of all the angles of a quadrilateral = 360°</em>
∠A + ∠B + ∠C + ∠D = 360°
∠A + ∠C + 180° = 360° (using (3))
∠A + ∠C = 360° – 180°
∠A + ∠C = 180° – – – – (4)
Equation (3) and (4) shows that,
Opposite angles of a cyclic quadrilateral are supplementary.
Hence proved.
