Question

Inscribed angle theorem

Answer 1

Answer:

The measure of an inscribed angle is half of the measure of the arc intercepted

by it.

67

Proof  ः  In D AOC,

side OA @ side OC ...... radii of the same circle.

\ Ð OAC = Ð OCA ..... theorem of isosceles triangle.

Let Ð OAC = Ð OCA = x ...... (I)

Now, Ð EOC = Ð OAC + Ð OCA .... exterior angle theorem of a triangle.

= x° + x° = 2x°

But Ð EOC is a central angle.

\ m(arc EC) = 2x° .... definition of measure of an arc ..... (II)

\ from (I) and (II).

Ð OAC = Ð EAC =

1

2

m(arc EC) ..... (III)

Similarly, drawing seg OB, we can prove ÐEAB = 1

2

m(arc BE)...... (IV)

\ÐEAC + ÐEAB =

1

2

m(arc EC)+ 1

2

m(arc BE).... from (III) and (IV)

\ ÐBAC =

1

2 [m(arc EC) + m(arc BE)]

=

1

2 [m(arc BEC)] =

1

2 [m(arc BDC)] ..... (V)

Note that we have to consider three cases regarding the position of the centre of

the circle and the inscribed angle. The centre of the circle lies (i) on one of the arms of

the angle (ii) in the interior of the angle (iii) in the exterior of the angle. Out of these,

first two are proved in (III) and (V). We will prove now the third one.

In figure 3.46,

Ð BAC = Ð BAE - Ð CAE

=

1

2

m(arc BCE) - 1

2

m(arc CE)

...... from (III)

=

1

2 [m(arc BCE) - m(arc CE)]

=

1

2 [m(arc BC)] ...... (VI)

The above theorem can also be stated as follows.

The measure of an angle subtended by an arc at a point on the circle is half of

the measure of the angle subtended by the arc at the centre.