Question

prove that the angle substended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle

Answer 1

Given :
An arc ABC of a circle with center O , and a point C on the remaining part of the circumference.

To Prove :
Angle AOB = Twice angle ABC

Construction : Join OC and produce it to a suitable point D

Proof : Let angles 1,2,3,4,5,6 be as shown in figure

a) OA=OB=OC [RADII OF THE SAME CIRCLE ARE EQUAL]

b)1=4,2=3 [ANGLES OPPOSITE TO EQUAL SIDES ARE EQUAL]

c)5=4+1=Twice angle1 [EXTERIOR ANGLE PROPERTY]

d)6=2+3=Twice angle 2 [EXTERIOR ANGLE PROPERTY]

e)5+6=Twice angle (1+2) [ADD STATEMENT c AND d]

f)Angle AOB = Twice angle ACB [WHOLE = SUM OF ALL ITS PARTS]

hope it will help u......

thanks......!

Answer 2

Given. C(o,r)
To prove . Angle AOB = 2angle ACB
Construction. Join co and produced to p
Proof. In AOC
Angle 3 = angle 1+ angle 5
Angle 1= angle 5 (OC - OA)
Angle 3 = 2angle 1.......(1)
Similarly , In ∆BOC
Angle 4 = 2angle 2...(2)
Add equation (1) and (2)
We get angle 3 + angle 4 = 2(angle1+angle2)
Angle AOB = 2 angle ACB.
Hence proved...