CAT Angle Subtended by an Arc of a Circle - Practice Questions & MCQ

From the previous concepts we know that the congruent arcs (or equal arcs) of a circle subtend equal angles at the centre.

Therefore, the angle subtended by a chord of a circle at its centre is equal to the angle subtended by the corresponding (minor or major) arc at the centre.

Now let's study a theorem that gives the relationship between the angles subtended by an arc at the centre and at a point on the circle.

Theorem 8 : The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Let's prove this

Given a wircle C(O,r)in which arc PQ subtends ∠ POQ at the centre and ∠ PAQ at any point A on the remaining part of the circle.

We need to prove that ∠ POQ = 2 ∠ PAQ.

There are three different cases as given in the figure.In (i), arc PQ is minor; in (ii), arc PQ is a semicircle and in (iii), arc PQ is major.

Let us begin by joining AO and extending it to a point B.

In all the cases,

                      ∠ BOQ = ∠ OAQ + ∠ AQO

and,               ∠ BOP = ∠ OAP + ∠ APO

(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.)

Also in ∆ OAQ,

                          OA = OQ               (Radii of a circle)

Therefore,    ∠ OAQ = ∠ OQA

This gives    ∠ BOQ = 2 ∠ OAQ                        ...(1)

Again,                OA = OP               (Radii of a circle)

Therefore,    ∠ OAP = ∠ APO

This gives    ∠ BOP = 2 ∠ OAP                        ...(2) 

From (1) and (2),

     ∠ BOP + ∠ BOQ = 2(∠ OAP + ∠ OAQ)

This is the same as

                    ∠ POQ = 2 ∠ PAQ                         ...(3)

For the case (iii), where PQ is the major arc, (3) is replaced by

   reflex angle POQ = 2 ∠ PAQ.

Theorem 9 : Angles in the same segment of a circle are equal.

Suppose we join points P and Q and form a chord PQ in the above figures. Then ∠ PAQ is also called the angle formed in the segment PAQP.

We have to prove ∠ PCQ = ∠ PAQ.

From Theorem 8, A can be any point on the remaining part of the circle. So if we take any other point C on the remaining part of the circle, we have 

                       ∠ POQ = 2 ∠ PCQ = 2 ∠ PAQ

Therefore,               ∠ PCQ = ∠ PAQ.

Property 1: The angle in a semicircle is a right angle.

Consider the figure

Here ∠PAQ is an angle in the segment, which is a semicircle.

Also, 

(As POQ is a straight line)

If we take any other point C on the semicircle, again we will get ∠ PCQ = 90°.

 

Theorem 9 : Angles in the same segment of a circle are equal.

Suppose we join points P and Q and form a chord PQ in the above figures. Then ∠ PAQ is also called the angle formed in the segment PAQP.

We have to prove ∠ PCQ = ∠ PAQ.

From Theorem 8, A can be any point on the remaining part of the circle. So if we take any other point C on the remaining part of the circle, we have 

                       ∠ POQ = 2 ∠ PCQ = 2 ∠ PAQ

Therefore,               ∠ PCQ = ∠ PAQ.

Property 1: The angle in a semicircle is a right angle.

Consider the figure

Here ∠PAQ is an angle in the segment, which is a semicircle.

Also, 

(As POQ is a straight line)

If we take any other point C on the semicircle, again we will get ∠ PCQ = 90°.

 

 

Theorem 10 : If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).

You can see the truth of this result as follows:

AB is a line segment and C, D are two points lying on the same side of AB such that ∠ ACB = ∠ ADB.

We have to show that the points A, B, C and D lie on a circle. 

Let us draw a circle through the points A, C and B.

If D lies on the circle passing through A, B and C then clearly the result follows.

If possible, suppose D does not lie on this circle.

Then, this circle will intersect AD or AD produced in D'.

Now,                ∠ ACB = ∠ ADB             (given)

Now,               ∠ ACB = ∠ AD'B             ([angles in the same segment)

Therefore,      ∠ ADB = ∠ AD'B

But, an exterior angle of a triangle can never be equal to its interior opposite angle.

So, ∠ ADB = ∠ AD'B is true unless D' coincide with D.

Thus, D lies on the circle passing through A, B and C.

Hence, the points A, B, C, D are concyclic.