Question

prove that the angle subtended by an arc at the center is twice of the angle subtended by the same arc at 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 ABCConstruction : Join OC and produce it to a suitable point DProof : 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]

Answer 2

$\textsf{\huge{\underline{Hello Mate!}}}$

Given : PQ is a chord where (i) PQ is minor arc, (ii) PQ is semicircular arc and (iii) PQ is major arc.

To prove : < POQ = 2 < PRQ.

To construct : Extend OR passing through center of circle.

Proof : Please refer to attachment above.

Concept used : Exterior angle property whivh states that sum of two interior opposite angle is equal to one exterior angle.

Mathematically, ext. < = opp. [ < interior + < interior ]

$\textsf{\red{Hence proved.}}$

$\textbf{\large{Q.E.D}}$

$\textbf{Have great future ahead!}$