Question

can you help me please?

Answer 1

Heya User,

--> CD = OD = OC 
=> OCD is an equilateral triangle of measure = 60°

Now, Angle COD = 2Angle CBD = 60°  [ Center - segment ]
       => Angle CBD = 30°              ---------> ( i )


Also, Angle subtended by a diameter on circle = 90°
=> Angle ACB = 90°
=> Angle ( CBA + CAB ) = 90° ---> ( ii )

Add ( i ) and ( ii ) to get -->
---> Angles ( EBA + EAB ) = 90° + 30° = 120°

=> In triangle EBA, m( Angle AEB ) = 180° - 120° = 60°

Hence, Angle AEB = 60°

Answer 2

ΔODC is equilateral OC, OD and CD are radii)
∴ ∠COD = 60°

Now, ∠CBD = ¹/₂ ∠COD (the angle subtended by an arc is double the angle                                         subtended by it at any point on the remaining part of                                          the circle)
So, ∠CBD = 30°
Again, ∠ACB = 90° (same reason)
So, ∠BCE = 180° - ∠ACB
                 = 90°
In ΔCEB,
∠CBE + ∠BCE + ∠CEB = 180° (ASP)
∠CEB = 180° - (90° + 30°)
           = 180° - 120°
∠CEB = 60° =∠AEB

                                                  Hence, Proved