Question

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

Answer 1

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Answer 2

Answer: [The diagram is above in my friend's account ]

<AEB = 60

Step-by-step explanation:

ΔODC is Equilateral (The angles subtended by two chords at the centre are equal, then the chords are congruent)

∴ <COD = 60

Now,

= <CBD = 1/2 <COD (The pair of opposite angles of a cyclic quadrilateral are supplementary)

= <CBD = 30

= <ACB = 90 (Angle subtended by an arc at the centre of the circle is twice the angle subtended by it at any other point on the circle)

= <BCE = 180 - <ACB = 90 (Linear Pair)

= <BCE = 90 - 30 = 60, i.e. <AEB = 60

∴ <AEB = 60

Hence Proved