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The angle subtended by an arc at centre is double the angle subtended by it at any point on the remaining part of the circle Prove!

Sorry I forgot to join the figure please can u ask this question again please please
I edited the answer live :);)



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
              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]

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It is clearly seen that the angle at C is twice the angle at D. Here we have the fundamental theorem with regard to angles in circles -
" the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the circle " 

This can be proved as follows : 
Given a circle (C,r) in which arc AB subtends angle (ACB) at the centre and angle (ADB) at any point D on the circle. 

To Prove: 
(i) angle (ACB) = angle (ADB), when arc AB is a minor arc or a semicircle. 
(ii) Reflex angle (ACB)=2× angle (ADB), when arc AB is a major arc 

Construction: Join AB and DC. Produce DC to a point X outside the circle.

Clearly, there are 3 cases.
Case I : arc AB is a minor arc [Fig (i)] 
Case II : arc AB is a semicircle [Fig (ii)] 
Case III : arc AB is a major arc [Fig (iii)] 

We know that when one side of a triangle is produced then the exterior angle so formed is equal to the sum of the interior opposite angles. 

angle (ACX) = angle (CAD) + angle (CDA) [consider triangle CAD
angle (BCX) = angle (CBD) + angle (CDB) [consider triangle CBD

angle (CAD) = angle (CDA) [since CD=CA=r ] and , 
angle (CBD) = angle (CDB) [since CD=CB=r ] 

angle (ACX)=2× angle (CDA) and 
angle (BCX)=2×angle (CDB) 

In Fig (i) and Fig (ii) , 
angle (ACX) + angle (BCX) =  angle (CDA) +  angle (CDB)⇒ 
angle (ACB) = 2[angle (CDA) + angle (CDB)
angle (ACB) = angle (ADB) 

in Fig (iii), 
angle (ACX) + angle (BCX) =  angle (CDA) +  angle (CDB)⇒ 
reflex angle (ACB) = 2[angle (CDA) + angle (CDB)⇒ 
reflex angle (ACB) =  angle (ADB) 

Hence proved. 
This theorem has 2 main implications:

(i) The angle in a semicircle is a right angle .
i.e. in fig (2), where arc AB is a semicircle, or in other words ACB is a diameter, 
angle (ADB)=12× angle (ACB)=12×(180 degrees) = 90 degrees = a right angle

This holds true for any position of point C on the semicircle. 

(ii) Angles in the same segment of a circle are equal. 
This means that if an arc subtends 2 angles, at 2 different points on the circle, these angles will be equal. 

We can prove this, by proving that each of the 2 angles is equal to half the angle subtended at the centre.

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