Question

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

Answer 1

cnstruction.. join OC,OD,BC[radii]
therfore COD is an equilateral triangle
cod=60
cbd=1/2*cod=30
acb=900
bce=900[as it is a semicircle]
in bce
30+90+ceb=180
ceb=180-120=60
aeb=60

Answer 2

Join oc od and bc
triangle odc is equilateral
therefore, <cod = 60
now, < cbd =1\2<cod (theorem )
this gives <cbd=30
again, <acb=90
so, <bce=180-<acb=90
which gives <ceb=90-30=60,i. e., <aeb=60