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

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

Respuesta :

proof:

CD= radius r

OC=OD= radius

OCD is equilateral triangle

∠DCO=∠COD=∠ODC=60  degrees

∠ACB=90  degrees

 (Angle in semicircle)

∠DOC=2∠DBC (half angle)

∠DBC=30  degrees

∠ECB+∠BCA=180  degrees

 (linear pair)

∠ECB=180−90=90  degrees

In △ECB

∠CEB+∠ECB+∠CBE=180  degrees

∠CEB+90+30=180

∠CEB=60  degrees

*This is the full proof and ask me questions in comments if you have anything you don't understand! Also I would appreciate if you give me brainlliest!!