Question

AB is a diameter of circle with centre O. chord CD is equal to radius OC. AC and BD produced intersect at P. Prove that angle APB is 60°​

Answer 1

Answer:

Step-by-step explanation:

join bc and od

now

oc=od=cd

cod is an equilateral triangle

angle cod is 60° (1)

also

angle acb =90 (angle in semicircle ) (2)

w.k.t

angle substended by any arc on circles boundry is equal to the half of angle substended by same arc on circle

2angle bcd =angle cod

by equation1

angle cbd =30 (3)

now

angle cbd +angle apb =angle acb ( ext.angle =int.angle in triangle)

by equation2 and 3

30+angle apb =90

angle apb =60