Question

ABCD is a cyclic quadrilateral. If A = 80° and AC bisects BAD and BCD, calculate
(i) ABC
(ii) ADC.
Hence, prove that AC is a diameter of the circle.​

Answer 1

Answer:

If <A = 80° then the AC bisects the <A ..

therefore 80/2 = 40 the angle formed at both side of Side AC = 40° (equation-1)

if so the opposite angle of <A = 180-<A(cyclic Quadrilateral) = 180-80

ie,<BCD= 100°

AC also bisects the <BCD it cuts into two equal parts

that means 100/2= 50 (equation-2)

from equation 1 and 2 we can simply calculate the <ADC and <ABC.

<ABC = 180-(40+50)

<ADC = 180-(40+50)

so <ADC and <ABC are 90°

If we consider AC , the vertex is 90° (the angle opposite to diameter is 90° in a quadrilateral then it will on the circle)

if the perpendicular side are on the circle the side. AC will be its Diameter.