in figure O is the centre of the circle and a b is the diameter of semicircle ABCD if ab is equal to BC and angle AEC is equal to 50 degree find angle C B E angle C D E and Angle A O B prove that bo is parallel to Ce
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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
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