In the previous section we saw the following case:
• We are given a minor arc and the point P is marked on the major arc
In this section we will see the other case:
• We are given a major arc and the point P is marked on the minor arc
Consider fig.27.14 below:
1. Fig(a) shows the major arc AXB. It has a central angle co
2. It subtends an APB at a point P on the minor arc AYB
3. We have to prove that ∠APB = c⁄2.
4. Fig(b) shows the angles required for the calculations
5. The angles are marked in the same pattern that we saw in fig.27.8(a) in the previous section. Let us write them:
• Consider ΔAOP:
• OA = OP (∵ radii of the same circle)
• So ΔAOP is isosceles. Then base angles will be equal
• Let the base angles be equal to xo
• Then ∠PAO = ∠APO = xo. These are shown in cyan colour
• ∠AOP = (180 -2x)o (∵ sum of interior angles of a triangle is 180o) This angle is shown in magenta colour
6. Consider BOP:
• OB = OP (∵ radii of the same circle)
• So ΔBOP is isosceles. Then base angles will be equal
• Let the base angles be equal to yo
• Then ∠PBO = ∠BPO = yo. These are shown in white colour
• ∠ BOP = (180 -2y)o (∵ sum of interior angles of a triangle is 180o) This angle is shown in blue colour
7. Now consider the angles around the centre point O. There are three angles:
• ∠ AOP = (180 -2x)o
• ∠ BOP = (180 -2y)o
• ∠ AOB = co. This is the angle 'subtended by the chord AB' at the centre O
8. We know that, sum of these three angles must be equal to 360o. So we can write:
(180 -2x) + (180 -2y) + c = 360
⇒ 360 -2x -2y +c = 360 ⇒ c = 2x + 2y ⇒ c = 2(x + y) ⇒ (x + y) = c⁄2
9. But (x + y) is our required ∠APB. We find that, it is equal to c⁄2. Hence proved.
• We took a 'convenient' point P for the above calculations. We can take a 'not so convenient' point also. Such a point P is shown in fig(c). The procedure is the same. The reader is advised to write all the steps for fig.27.14(c) also.
10. So in this case also we can write the same four steps:
(i) We are given an arc AXB. It has a central angle of co
(ii) Draw the alternate arc AYB. Mark a point P any where on the alternate arc AYB
(iii) The original arc AXB will subtend an ∠APB at P
(iv) Where ever be the point P on the arc AYB, the ∠APB will be equal to c⁄2.
Let us see an example:
In the fig.27.15(a) above:
(i) We are given an arc AXB. It has a central angle of 240o
(ii) Draw the alternate arc AYB. Mark a point P any where on AYB
(iii) The original arc AXB will subtend an ∠APB at P
(iv) Where ever be the point P on the arc AYB, the ∠APB will be equal to 240⁄2 = 120o.
■ Figs. (b) and (c) shows that P can be any where on the arc AXB. The angle 120o will not change.
11. The above fact can be used to calculate the angle at P. Let us see some examples:
In fig.27.16(a) above:
(i) We are given an arc AB (magenta coloured). It has a central angle of 160o
(ii) A point P is marked on the alternate arc.
• What is the angle subtended at P, by the magenta coloured arc?
Solution:
The required angle will be equal to 160⁄2 = 80o
In fig.27.16(b) above:
(i) We are given an arc AB (magenta coloured). It has a central angle of 108o
(ii) A point P is marked on the alternate arc.
• What is the angle subtended at P, by the magenta coloured arc?
Solution:
The required angle will be equal to 108⁄2 = 54o
In fig.27.16(c) above:
(i) We are given an arc AB (magenta coloured). It has a central angle of 70o
(ii) A point P is marked on the alternate arc.
• What is the angle subtended at P, by the magenta coloured arc?
Solution:
The required angle will be equal to 70⁄2 = 35o.
1. Consider any arc. Let it have a central angle of co.
2. Let P be any point on the alternate arc
3. The original arc will subtend an angle at P
4. This angle will be equal to c⁄2
2. In fig(a), an arc AYB has a central angle of co.
• P is a point on the alternate arc AXB
• The arc AYB subtends ∠APB at P
• We know that ∠APB will be equal to c⁄2.
3. Now, keeping all other points the same, we move the point P along the arc AXB. The direction of motion is indicated by the white arrow
4. Since A and B remains at the same positions, the ∠APB will remain c⁄2 even when we move P
5. After some time, the line AP becomes closer and closer to OA. This is shown in fig(b)
6. When we continue moving P, a stage will reach when AP completely cover OA. This is shown in fig(c)
7. In such situations, we must do careful analysis to distinguish between the central ∠AOB and the subtended ∠APB
• We are given a minor arc and the point P is marked on the major arc
In this section we will see the other case:
• We are given a major arc and the point P is marked on the minor arc
Consider fig.27.14 below:
 |
| Fig.27.14 |
2. It subtends an APB at a point P on the minor arc AYB
3. We have to prove that ∠APB = c⁄2.
4. Fig(b) shows the angles required for the calculations
5. The angles are marked in the same pattern that we saw in fig.27.8(a) in the previous section. Let us write them:
• Consider ΔAOP:
• OA = OP (∵ radii of the same circle)
• So ΔAOP is isosceles. Then base angles will be equal
• Let the base angles be equal to xo
• Then ∠PAO = ∠APO = xo. These are shown in cyan colour
• ∠AOP = (180 -2x)o (∵ sum of interior angles of a triangle is 180o) This angle is shown in magenta colour
6. Consider BOP:
• OB = OP (∵ radii of the same circle)
• So ΔBOP is isosceles. Then base angles will be equal
• Let the base angles be equal to yo
• Then ∠PBO = ∠BPO = yo. These are shown in white colour
• ∠ BOP = (180 -2y)o (∵ sum of interior angles of a triangle is 180o) This angle is shown in blue colour
7. Now consider the angles around the centre point O. There are three angles:
• ∠ AOP = (180 -2x)o
• ∠ BOP = (180 -2y)o
• ∠ AOB = co. This is the angle 'subtended by the chord AB' at the centre O
8. We know that, sum of these three angles must be equal to 360o. So we can write:
(180 -2x) + (180 -2y) + c = 360
⇒ 360 -2x -2y +c = 360 ⇒ c = 2x + 2y ⇒ c = 2(x + y) ⇒ (x + y) = c⁄2
9. But (x + y) is our required ∠APB. We find that, it is equal to c⁄2. Hence proved.
• We took a 'convenient' point P for the above calculations. We can take a 'not so convenient' point also. Such a point P is shown in fig(c). The procedure is the same. The reader is advised to write all the steps for fig.27.14(c) also.
10. So in this case also we can write the same four steps:
(i) We are given an arc AXB. It has a central angle of co
(ii) Draw the alternate arc AYB. Mark a point P any where on the alternate arc AYB
(iii) The original arc AXB will subtend an ∠APB at P
(iv) Where ever be the point P on the arc AYB, the ∠APB will be equal to c⁄2.
Let us see an example:
 |
| Fig.27.15 |
(i) We are given an arc AXB. It has a central angle of 240o
(ii) Draw the alternate arc AYB. Mark a point P any where on AYB
(iii) The original arc AXB will subtend an ∠APB at P
(iv) Where ever be the point P on the arc AYB, the ∠APB will be equal to 240⁄2 = 120o.
■ Figs. (b) and (c) shows that P can be any where on the arc AXB. The angle 120o will not change.
11. The above fact can be used to calculate the angle at P. Let us see some examples:
 |
| Fig.27.16 |
(i) We are given an arc AB (magenta coloured). It has a central angle of 160o
(ii) A point P is marked on the alternate arc.
• What is the angle subtended at P, by the magenta coloured arc?
Solution:
The required angle will be equal to 160⁄2 = 80o
In fig.27.16(b) above:
(i) We are given an arc AB (magenta coloured). It has a central angle of 108o
(ii) A point P is marked on the alternate arc.
• What is the angle subtended at P, by the magenta coloured arc?
Solution:
The required angle will be equal to 108⁄2 = 54o
In fig.27.16(c) above:
(i) We are given an arc AB (magenta coloured). It has a central angle of 70o
(ii) A point P is marked on the alternate arc.
• What is the angle subtended at P, by the magenta coloured arc?
Solution:
The required angle will be equal to 70⁄2 = 35o.
We considered both major arcs and minor arcs. The four steps will work for both the arcs. So in general we can apply it to any arc. The four steps are essential to solve many problems in science and engineering. We will write them as a theorem. We will write it in steps:
Theorem 27.4:1. Consider any arc. Let it have a central angle of co.
2. Let P be any point on the alternate arc
3. The original arc will subtend an angle at P
4. This angle will be equal to c⁄2
■ We can write the converse also:
1. Consider any arc.
2. Let P be any point on the alternate arc
3. Let the angle subtended at P, by the original arc be ao.
4. Then the central angle co of the original arc is given by: co = 2 × ao
1. Consider fig.27.17 below:1. Consider any arc.
2. Let P be any point on the alternate arc
3. Let the angle subtended at P, by the original arc be ao.
4. Then the central angle co of the original arc is given by: co = 2 × ao
A very interesting case:
 |
| Fig.27.17 |
• P is a point on the alternate arc AXB
• The arc AYB subtends ∠APB at P
• We know that ∠APB will be equal to c⁄2.
3. Now, keeping all other points the same, we move the point P along the arc AXB. The direction of motion is indicated by the white arrow
4. Since A and B remains at the same positions, the ∠APB will remain c⁄2 even when we move P
5. After some time, the line AP becomes closer and closer to OA. This is shown in fig(b)
6. When we continue moving P, a stage will reach when AP completely cover OA. This is shown in fig(c)
7. In such situations, we must do careful analysis to distinguish between the central ∠AOB and the subtended ∠APB
Now we will see some solved examples
Solved example 27.2:
All the four vertices of a rectangle are on a circle. Prove that the diagonal of the rectangle will be a diameter of the circle.
Solution:
1. It is better to first draw a rough sketch for this problem. Fig.27.18(a) below is a rough sketch
• ABCD is a quadrilateral with all it's vertices on the rectangle
• Given that ABCD is a rectangle. In the rough sketch, the angles at the vertices need not be exactly 90o.
2. Separate out an arc ABC. This is shown in fig(b).
3. This arc ABC subtends an ∠ADC = 90o, at a point D on the alternate arc
4. So, using the converse of theorem 27.4, the central ∠AOC of arc ABC will be ( 2 × 90o) = 180o.
5. If ∠AOC is 180o, the points A, O and C will lie on a straight line.
6. A and C lies on the circle, and O is the centre. If all those three points lie on a straight line, AOC will be a diameter. Hence proved.
7. An actual construction is shown in fig(c)
8. Using the same method, we can prove that the other diagonal will also be a diameter of the circle pass through the centre of the circle. The reader may write the proof for that also.
The above findings can be used as a standard result. Let us write the steps:
Solved example 27.2:
All the four vertices of a rectangle are on a circle. Prove that the diagonal of the rectangle will be a diameter of the circle.
Solution:
1. It is better to first draw a rough sketch for this problem. Fig.27.18(a) below is a rough sketch
• ABCD is a quadrilateral with all it's vertices on the rectangle
• Given that ABCD is a rectangle. In the rough sketch, the angles at the vertices need not be exactly 90o.
 |
| Fig.27.18 |
3. This arc ABC subtends an ∠ADC = 90o, at a point D on the alternate arc
4. So, using the converse of theorem 27.4, the central ∠AOC of arc ABC will be ( 2 × 90o) = 180o.
5. If ∠AOC is 180o, the points A, O and C will lie on a straight line.
6. A and C lies on the circle, and O is the centre. If all those three points lie on a straight line, AOC will be a diameter. Hence proved.
7. An actual construction is shown in fig(c)
8. Using the same method, we can prove that the other diagonal will also be a diameter of the circle pass through the centre of the circle. The reader may write the proof for that also.
1. Consider any rectangle in which, all the four corners lie on a circle
2. Then both the diagonals of that rectangle will be diameters of the circle
3. So the centre of the circle will bisect both the diagonals.
3. So the centre of the circle will bisect both the diagonals.
In the next section, we will see a few more solved examples.
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