AB is a diameter of the circle, CD is a chord equal to the radius of the circle. AC and BD when extented intersect at a point E. Prove that angle AEB = 60°
1 answer:
proof:
CD= radius r
OC=OD= radius
OCD is equilateral triangle
∠DCO=∠COD=∠ODC=60 degrees
∠ACB=90 degrees
(Angle in semicircle)
∠DOC=2∠DBC (half angle)
∠DBC=30 degrees
∠ECB+∠BCA=180 degrees
(linear pair)
∠ECB=180−90=90 degrees
In △ECB
∠CEB+∠ECB+∠CBE=180 degrees
∠CEB+90+30=180
∠CEB=60 degrees
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