Question

In the fig., 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 point E. Prove that angle AEB = 60 ​

Answer 1

Answer:

Given: AB is a diameter to circle

CD = r

To prove : AEB = 60

Construction: Join OC,OD,BC

Proof

Consider triangle COD where

OC = OD = CD = r

therefore, triangle COD is an equilateral triangle

COD = 60

CBD = 1/2 × COD

= 1/2 × 60

= 30

ACB = 90 ( Angle inscribed in a semi circle)

ECB = 90 ( Linear pair)

Consider triangle CEB

CEB + ECB + CBD = 180

CEB + 90 + 30 = 180

CEB + 120 = 180

CEB = 60

therefore , AEB = 60