Question

In figure AB and CD arc two chords of circle, intersecting at point E. Prove that ∠AEC = (Angle formed by arc CXA at center of angle formed by arc DYB at center)

Answer 1

O is centre of circle and arc is AC

∠AOC is angle subtended by arc AC at center, is double the angle ∠ABC, angle at remaining part by same arc

∴ ∠AOC = 2∠ABC

Similarly

∠BOD, angle subtended by are BD at center of circle is double the angle at remaining part.

∠BOD = 2∠BCD …..(ii)

Adding equation (i) and (ii)

∠AOC + ∠BOD = 2(∠ABC + ∠BCD)

⇒ ∠AOC + ∠BOD = 2∠AEC

⇒ ∠AEC = \(\frac { 1 }{ 2 }\) [∠AOC + ∠BOD]

⇒ ∠AEC = \(\frac { 1 }{ 2 }\) [Angle formed at center by CXA angle formed of center by chord DYB]