Question

Prove the Cyclic Quadrilateral Conjecture

Answer 1

Given ABCD is a cyclic quadrilateral.

To prove that ∠A + ∠C = 180° and ∠B + ∠D = 180°

Construction: Join AC and BD.

Proof:

Angles in the same segment are equal.

∠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 sum of the angles of a triangle is 180°.

In ΔABC, ∠ACD + ∠DAC + ∠ADC = 180°

Now, 180° = ∠ABC + ∠ADC

180° = ∠B + ∠C – – – – (3)

Sum of all the angles of a quadrilateral = 360°

∠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.