Question

in figure, O is the centre of the circle, SE is the diameter of the semicircle ABCDE. If AB=BC and <AEC = 50°, then find (a) <CBE (b) <CDE (c) <AOB​

Answer 1

your answer:

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

HOPE U UNDERSTOOD!!

#FAHEEM

Answer 2

Step-by-step explanation:

Let angle CBE = x

Angle CDE = y

And angle AOB = z

Construction = join BO & CO

In triangle ABE

angle E = 50

angle ABE = 90     ( semicircle angle )

By ASP angle A = 40

In triangle ABO

AO = BO so angle A = angle ABO = 40

By ASP 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 ------------> first equation

In first equation

40 = x + 50

x = -10

but we take it as +10 so in cyclic quadrilateral BCDE x + y = 180

10 + y = 180

y = 170

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