Question -
Answer -
In ΔABC,
∠ABC + ∠BCA + ∠CAB = 180°(Angle sum property of a triangle)
⇒ 90° + ∠BCA + ∠CAB = 180°
⇒ ∠BCA + ∠CAB = 90° … (1)
In ΔADC,
∠CDA + ∠ACD + ∠DAC = 180°(Angle sum property of a triangle)
⇒ 90° + ∠ACD + ∠DAC = 180°
⇒ ∠ACD + ∠DAC = 90° … (2)
Adding equations (1) and (2), we obtain
∠BCA + ∠CAB + ∠ACD + ∠DAC = 180°
⇒ (∠BCA + ∠ACD) + (∠CAB + ∠DAC) = 180°
∠BCD + ∠DAB = 180° … (3)
However, it is given that
∠B + ∠D = 90° + 90° = 180° … (4)
From equations (3) and (4), it can be observed that thesum of the measures of opposite angles of quadrilateral ABCD is 180°.Therefore, it is a cyclic quadrilateral.
Consider chord CD.
∠CAD = ∠CBD (Angles in the same segment)
