Question

Example 3 : 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 ZAEB = 60°. E od BO:​

Answer 1

Answer:

CD= radius r

OC=OD= radius

OCD is equilateral triangle

∠DCO=∠COD=∠ODC=60o

∠ACB=90o (Angle in semicircle)

∠DOC=2∠DBC (half angle)

∠DBC=30o

∠ECB+∠BCA=180o (linear pair)

∠ECB=180−90=90o

In △ECB

∠CEB+∠ECB+∠CBE=180o

∠CEB+90+30=180

∠CEB=60o