Question

in the adjoining figure,O is the centre and AOE is the diameter of the semicircle ABCDE.If AB=BC and ANGLEAEB=50,find angle CBE angle CDE angle AOB

Answer 1

Let angle CBE =x

Angle CDE =y

And angle AOB=z

Construction =join BO & CO

Proof :

>in triangle ABE

angle E= 50

angle ABE =90

(semicircle angle)

By angle sum property angleA =40

>In triangle ABO

AO=BO so angle A=angle ABO =40

By angle sum property z=100

In triangle ABC & BOC

AO=OC (radius of circle)

BO=BO (common)

AB=BC (given)

By SSS triangle ABO is congruent to triangle BOC

By cpct :angle BAO=BCO

angle AOB =BOC

angle ABO=CBO... 1 st equation

In 1st equation

40=x+50

x=-10 or 10 so in cyclic quadrilateral BCDE x+y=180

10+y=180

y =170

Answer 2

angel in the drawing is given as angel AEC=50°