Question

in fig 10.32,AB is a diameter of the circle,CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E.Prove that angle AEB=60°​

Answer 1

Step by step

Join OC,OD and BC

In △OCD, we have

OC = OD = CD (Each equal to radius) ∴ △ODC is an equilateral triangle. ⇒ ∠COD = 600

Also, ∠COD = 2∠CBD

⇒ 600 = 2∠CBD ⇒ ∠CBD = 300

Since ∠ACB is angle in a semi-circle.

∠ACB = 900

⇒ ∠BCE = 1800 - ∠ACB = 1800 - 900 = 900

Thus, in △BCE,we have

∠BCE = 900 and ∠CBE = 300

∴ ∠BCE + ∠CEB + ∠CBE = 1800

⇒ 900 + ∠CEB + 300 = 180 ⇒ ∠CEB = 600

Hence, ∠AEB = ∠CEB = 600

Answer 2

Answer:

I think it will give u a better explanation