Answer:
Step-by-step explanation:
Measure of an inscribed angle intercepted by an arc is half of the measure of the arc.
From the picture attached,
m(∠A) = [tex]\frac{1}{2}m(\text{arc BD})[/tex]
= [tex]\frac{1}{2}[m(\text{BC})+m(\text{CD}][/tex]
= [tex]\frac{1}{2}[55^{\circ}+145^{\circ}][/tex]
= 100°
m(∠C) = [tex]\frac{1}{2}[(360^{\circ})-m(\text{arc BCD})][/tex]
= [tex]\frac{1}{2}(360^{\circ}-200^{\circ})[/tex]
= 80°
m(∠B) + m(∠D) = 180° [ABCD is cyclic quadrilateral]
115° + m(∠D) = 180°
m(∠D) = 65°
m(arc AC) = 2[m(∠D)]
m(arc AB) + m(arc BC) = 2(65°) [Since, m(arc AC) = m(arc AB) + m(arc BC)]
m(arc AB) + 55° = 130°
m(arc AB) = 75°
m(arc ADC) = 2(m∠B)
m(arc AD) + m(arc DC) = 2(115°)
m(arc AD) + 145° = 230°
m(arc AD) = 85°
