Chapter 27.5 - Angle at a point Inside or Outside the circle

In the previous sections we saw theorem 27.4 and it's converse. We also saw some solved examples.
■ So far we have been considering the following case:
• An arc subtends an angle at a point P on the alternate arc.
    ♦ That means, the point P is situated some where on the circle 
■ Now we consider the other case:
• An arc subtends an angle at a point P which is not on the circle.
• Such a case is shown in the fig.27.32(a) below: 

Fig.27.32

• In this case, P is in the interior of the circle. We want to know the ∠APB
For that, the following steps are used:
1. Extend PA along the same line, to meet the circle at M
2. Extend PB along the same line, to meet the circle at N
3. Name arc AB as AXB and arc MN as MYN. This is shown in fig(b)
4. Let the central angle of arc AXB be x^o. This is shown in fig(c)
5. So, if this arc AXB subtends an angle at any point on it's alternate arc, that angle would be ^x⁄₂.
Is there any such point?
Indeed there is.
6. If we join A to N, we will see that, arc AXB subtends an ∠ANB at the point N on the alternate arc.
So ∠ANB = ^x⁄₂. This is shown in fig(d) below:

Fig.27.32

7. Let the central angle of arc MYN be y^o
8. So, if this arc MYN subtends an angle at any point on it's alternate arc, that angle would be ^y⁄₂
Is there any such point?
Indeed there is.
9. Arc MYN subtends an ∠MAN at the point A on the alternate arc.
So ∠MAN = ^y⁄₂
10. Now consider ΔPAN in fig(e). 
• ∠ APB is an exterior angle in PAN
• Exterior angle = sum of remote interior angles.
11. So we get: ∠APB = ^x⁄₂ + ^y⁄₂ ⇒ APB = ^(x+y)⁄₂

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We can write the above result in the form of a theorem. We will write it in steps:

Theorem 27.5

1. Consider any arc AXB
2. Let it subtend an ∠APB at a point P in the interior of the circle
3. Also let the extensions of the two legs AP and BP subtend an arc MYN on the circle
4. Then ∠APB is the average of the following two items:
(i) Central angle of arc AXB
(ii) Central angle of arc MYN

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Another case is shown in the fig.27.33(a) below:

Fig.27.33

• In this case, P is in the exterior of the circle. We want to know the ∠APB

For that, the following steps are used:
1. Let the leg AP intersect the circle at M
2. Let the leg BP intersect the circle at N
3. Name arc AB as AXB and arc MN as MYN. This is shown in fig(b)
4. Let the central angle of arc AXB be x^o. This is shown in fig(c)
5. So, if this arc AXB subtends an angle at any point on it's alternate arc, that angle would be ^x⁄₂
Is there any such point?
Indeed there is.
6. If we join A to N, we will see that, arc AXB subtends an ∠ANB at the point N on the alternate arc.
So ANB = ^x⁄₂. This is shown in fig(d)

Fig.27.33

7. Let the central angle of arc MYN be y
8. So, if this arc MYN subtends an angle at any point on it's alternate arc, that angle would be ^y⁄₂
Is there any such point?
Indeed there is.
9. Arc MYN subtends an ∠MAN at the point A on the alternate arc.
So ∠MAN = ^y⁄₂
10. Now consider PAN in fig(e). 
• ∠ ANB is an exterior angle in ΔPAN
• Exterior angle = sum of remote interior angles.
11. So we get: ∠ANB = ∠APB + ∠PAN ⇒^x⁄₂ = ∠APB + ^x⁄₂ ⇒ ∠APB = ^(x-y)⁄₂

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We can write the above result in the form of a theorem. We will write it in steps:

Theorem 27.6

1. Consider any arc AXB
2. Let it subtend ∠APB at a point P in the exterior of the circle
3. Also let the two legs AP and BP intersect the circle at M and N respectively
4. Then ∠APB is calculated as follows:
(i) Write the central angle of arc AXB
(ii) Write the central angle of arc MYN
(iii) Subtract the smaller from the larger
(iv) Half of the result of subtraction will be ∠APB

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Now we will see some solved examples

Solved example 27.12

In the fig.27.34(a) below, central angle of arc AXB is 40^o. Central angle of arc CYD is 70^o. 

Fig.27.34

Find the angles of ΔAPD.

Solution:

• P is an interior point. The arc AXB makes ∠APB at P

• The legs AP and BP are extended to meet the circle at C and D respectively.

• So we can use theorem 27.5 to calculate ∠APB. But in this problem we are not asked to find ∠APB
• We are asked to find the interior angles of ΔAPD. So direct application of theorem 27.5 is not possible

• However we need to apply some results that we came across while deriving theorem 27.5. The steps are given below:

1. Consider fig.27.34(b). 

• Arc AXB and arc AYB are alternate arcs

• Arc AXB subtends ∠ADB on the alternate arc

• Arc AXB has a central angle of 40^o. So ∠ADB = ¹⁄₂ × 40 = 20^o

• But ∠ADB = ∠ADP. So we got the interior angle of ΔAPD at D  

2. Consider fig.27.34(c). 

• Arc CYD and arc CXD are alternate arcs

• Arc CYD subtends ∠DAC on the alternate arc

• Arc CYD has a central angle of 70^o. So ∠DAC = ¹⁄₂ × 70 = 35^o

• But ∠DAC = ∠DAP. So we got the second interior angle of ΔAPD at P

3. Now the third interior angle, which is at P = [180-(20+35)] = [180-55] = 125^o. 

4. So the required interior angles are:

∠D = 20^o, ∠A = 35^o and ∠P = 125^o.

Solved example 27.13
In fig.27.35(a) below, AB is a diameter of the circle. A, P, B and R are four points on the circle. Lines AP and RB intersect at Q. Find ∠PRB, ∠PBR and ∠BPR 

Fig.27.35

Solution:
1. Separate out arc PYB as shown in fig(b). It is subtending ∠BAP = 34^o on the alternate arc PXB
2. This same arc PYB is subtending another ∠PRB also on the alternate arc PXB
So ∠BAP = ∠PRB = 34^o.
3. Separate out arc AXR. It is subtending ∠ABR on the alternate arc AYR  
So ∠ABR will be half of the central angle x of the arc AXR. That is., ∠ABR = ^x⁄₂.
4. This ∠ABR is an exterior angle of ΔABQ
• Exterior angle = sum of remote interior angles 
• ^x⁄₂ = 34 + 28 ⇒ ^x⁄₂ = 62
5. So from (3) above, we get: ∠ABR = ^x⁄₂ = 62
6. The arc AXR is subtending ∠ABR on the alternate arc AYR
• The same arc AXR is subtending ∠APR on the alternate arc AYR
• So ∠ABR = ∠APR = 62^o
7. Now consider ΔPRB. We are asked to find all it's interior angles
We have already obtained ∠PRB as 34^o in (2)
8. AB is a diameter. So ∠APB = 90^o. (Theorem 27.1)
∠BPR = ∠APB - ∠APR ⇒ ∠BPR = 90 - 62 ⇒ ∠BPR = 28^o.
9. The remaining ∠PBR = [180 -(34+28)] = [180-62] = 118^o.
10. So the required angles are: 
∠PRB = 34^o, ∠PBR = 118^o and ∠BPR = 28^o.

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In the next section, we will see Segments of a circle.

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Chapter 27.4 - Angles subtended by Arcs - Solved examples - 2

In the previous sections we saw theorem 27.4 and it's converse. We also saw some solved examples. In this section we will see a few more solved examples.

Solved example 27.8
How do we draw a 22¹⁄₂^o angle?
Solution:
• Let us first see what happens when we double a 22¹⁄₂^o angle. Fig.27.26 below shows a rough sketch
• In fig(a), an arc AB subtends an angle of 22¹⁄₂^o at P. 
• Separate out this arc AB as shown in fig(b). We see that, the central angle of arc AB is (22.5 × 2) = 45^o.
• Now we want to double this 45^o. So we want this 45^o in another circle. For that, with any convenient point O₁ as centre and OO₁ as radius, draw a circle. This is shown in fig(c)

Fig.27.26

• Since the centre is O₁, and radius OO₁, this circle will pass through O. So we have an angle of 45^o in the new circle. 
• This angle will cut an arc CD in the circle, and that arc CD will have a central ∠CO_1D of 45 × 2 = 90^o 
• This 90^o is shown in cyan colour in fig(c)
• So we made a 90^o from a 22¹⁄₂^o.
■ We can do the reverse. That is., we can make a 22¹⁄₂^o from a 90^o. The steps are as follows:
1. Draw a circle with centre O and also draw a diameter AB. This AB should preferably in the vertical direction, so that, our drawing will proceed towards the right in the horizontal direction. This is shown as green dashed line in fig.27.27(step 1) below. 
2. The diameter AB splits the circle into two semi-circles. Mark a point O₁ any where on the right semi-circle. Draw AO₁ and BO₁. By theorem 27.1, ∠AO₁B will be 90^o. This is shown in cyan colour in step 1. The point O₁ should be approximately on a horizontal line through O. Then the construction will proceed towards the right in a horizontal direction.

Fig.27.27

3. Draw a circle with centre O₁ and with any convenient radius so that, it will intersect both the legs of ∠AO₁B. at points C and D. This is the red coloured circle in fig.27.27(step 2)
• So the arc CD have a central ∠CO₁D of 90^o.
4. Mark a point O₂ on the red circle, approximately in line with O and O₁. The angle subtended at O₂ by the arc CD will be equal to (¹⁄₂×90) = 45^o.
• So we halved 90^o. We can continue like this and obtain the half of 45^o.
5. Draw a circle with centre O₂ and with any convenient radius so that, it will intersect both the legs of ∠CO₂D. at points E and F. This is the green coloured circle in fig.27.27(step 3)
• So the arc EF has a central ∠EO₂F of 45^o.
6. Mark a point P on the green circle, approximately in line with O, O₁ and O₂. The angle subtended at P by the arc EF will be equal to (¹⁄₂×45) = 22¹⁄₂^o.
■ So we finally obtained 22¹⁄₂^o.    

Solved example 27.9
In the fig,27.28(a) below, O is the centre of the circle and line OC is parallel to line PB. 

Fig.27.28

(i) Prove that OC bisects ∠AOB
(ii) Explain how this can be used to draw the bisector of an angle
Solution:
Part (i): 
1. Separate out the arc AB as shown in fig.27.28(b). This arc AB has a central ∠AOB of c^o.
2. Also, this arc AB subtends ∠APB at P on the alternate arc. Then by theorem 27.4, ∠APB = ^c⁄₂^o.
3. Given that OC is parallel to PB. 
Consider fig(c):
• PB and OC are two parallel lines
• They are cut by a transversal PA
• So ∠PBA and ∠COA are corresponding angles and are equal
• Thus we get ∠COA = ^c⁄₂^o. That means, the line OC bisects ∠AOB
Part (ii):
1. Consider ∠AOB in fig.27.29(a) below. We want to draw the bisector of this angle

Fig.27.29

2. Draw a circle with centre O
3. Extend the line OA towards the left so that it meets the circle at P. This is shown in fig(b)
4. Draw PB
5. Draw OC parallel to PB. Then OC is the bisector of ∠AOB
■ Note that this problem is related to the special case that we saw before

Solved example 27.10
In the fig.27.30(a) below, O is the centre of the circle. Prove that x + y = 90^o.

Fig.27.30

Solution:
1. Consider ΔAOB. The sides OA and OB are equal. Because they are radii of the same circle.
2. So ΔOAB is an isosceles triangle. The base angles will be equal. Thus ∠OAB = ∠OBA = x^o. This is shown in fig(b)
3. So we get ∠AOB = [180 - (x+x)] = [180-2x]
4. Separate out the arc ADB. The ∠AOB is the central angle of arc ADB. 
5. This arc ADB subtends an ∠ACB on the alternate arc
6. But ∠ACB is given as y^o.
7. So we can write: ∠ACB = y^o = half of [180-2x]
That is: y = ^[180-2x]⁄₂ ⇒ y = (90-x) ⇒ (x+y) = 90^o.

Solved example 27.11
In fig.27.31(a) below, PQ and RS are two mutually perpendicular chords of a circle.

Fig.27.31

∠QPR = 50^o. Find ∠PQS
Solution:
1. Separate out the arc PR as shown in fig(b).
2. This arc PR subtends ∠PRS on the alternate arc
3. This same arc PR subtends another ∠PQS on the alternate angle
4. So ∠PRS = ∠PQS
5. We are asked to find ∠PQS. So if we can find ∠PRS, we get the answer.
6. Let the chords PQ and RS intersect at T  
7. Consider ⊿PTR. It is given that PQ and RS are mutually perpendicular. So ∠PTR = 90^o.
8. ∠PRS = [180-(90+50)] ⇒ ∠PRS = [180-140] ⇒ ∠PRS = 40^o
9. Thus from (4) we get: ∠PQS = 40^o.

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A solved example is shown in the form of a video presentation here:
Solved example video presentation

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In the next section, we will see the cases when an arc subtends an angle at a point P which is not on the circle.

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Chapter 27.3 - Angles subtended by Arcs - Solved examples

In the previous section we saw theorem 27.4 and it's converse. We also saw a solved example. In this section we will see a few more solved examples.

Solved example 27.3
How do we draw a triangle with angles 40^o, 60^o and 80^o within a circle of 2.5 cm radius?
Solution:
1. First step is to draw a rough sketch. This is shown in fig.27.19(a) below.
• Let A, B and C be the three vertices. All of them lie on a circle

Fig.27.19

2. Join A and B to the centre O
3. Separate out the minor arc AB. This is shown in fig(b)
4. Now we have an arc AB which subtends ∠ACB on the alternate arc.
5. ∠ACB = 80^o. So, using the converse of theorem 27.4, we get: ∠AOB = 2 × 80 = 160. This is marked in fig(b)
6. Again separate out the minor arc BC. This is shown in fig(c)
7. Now we have an arc BC which subtends ∠BAC on the alternate arc.
8. ∠BAC = 40^o. So, using the converse of theorem 27.4, we get: ∠BOC = 2 × 40 = 80. This is marked in fig(c)
9. So now we have the angles between the radial lines from centre O. We can do the actual construction. 
10. At any convenient point O, draw a circle of radius 2.5 cm
11. Draw a radial line OA in any convenient direction. Draw OB such that ∠OAB is 160^o. This is shown in fig.27.20 below:        

Fig.27.20

11. Next draw OC such that ∠BOC is 80^o.
12. Thus the three vertices are obtained. Draw AB, BC and CA to get the required ΔABC.

Solved example 27.4
In the fig 27.21(a) below, what percentage of the total circumference is the length of the arc ADB

Fig.27.21

Solution:
1. Separate out the arc ADB as shown in fig.27.21(b)
2. This arc ADB subtends ∠ACB at point B on the alternate arc
3. ∠ACB = 60^o. So, using the converse of theorem 27.4, we get: ∠AOB = 2 × 60 = 120. This is marked in fig(b)
4. So we have the central angle of ADB. If we assume the radius of the circle to be 'r' cm, we can find the length of arc ADB. We can use theorem 21.1  
5. So length of arc ADB = ^πr⁄₁₈₀ ×120 = ^2πr⁄₃ . .
6. The whole circumference = 2πr
7. Now we take the ratio:
 ^{Length of arc ADB}⁄_{Whole circumference}  = ^2πr⁄₃ ÷ 2πr = ^2πr⁄₃ × ¹⁄_2πr = ¹⁄₃ = 33.33%
8. So length of the arc ADB is 33.33% of the whole circumference

Solved example 27.5
What is the radius of the circle shown in fig.27.22(a) below

Fig.27.22

Solution:
1. Separate out the arc ABC. This is shown in fig(b)
2. Consider ⊿ABC. Vertices A and C are on the circle. Angle at B is given as 90^o. 
• So by theorem 27.1, Arc ABC will be a semicircle. 
• That means, AC is the diameter. So midpoint of AC will be the centre O of the circle
• If we find the length of OA or OC, we will have the radius  
• So our aim is to find OA or OC. For that, the following steps can be used:
3. Angle at A = [180 - (90+45)] = [180 -90 -45] = 45^o.
4. So the base angles of ⊿ABC are equal. It is an isosceles triangle
Thus we get: AB = BC = 3 cm
5. So we have the two legs of ⊿ABC. Applying Pythagoras theorem, we get:
AC² = AB² + BC² ⇒ AC² = 3² + 3² ⇒ AC² = 2 × 3² ⇒ AC² = 2 × 9 ⇒ AC = 3√2
6. So OA = OC = ^AC⁄₂ = ^3√2⁄₂ = 1.5√2

Solved example 27.6
What is the area of the circle shown in fig.27.23(a) below

Fig.27.23

Solution:
1. We can use a standard result that we derived in the previous section. let us write it's steps:

(i) Consider any rectangle in which, all the four corners lie on a circle

(ii) Then both the diagonals of that rectangle will be diameters of the circle
(iii) So the centre of the circle will bisect both the diagonals. 

2. So the diagonal AC will be a diameter of the circle
3. AC will split the rectangle into two right angled triangles: ⊿ABC and ⊿ADC
4. Consider ⊿ADC. Hypotenuse is not given. 
• But one leg is 3 cm and the other leg is 4 cm. So it is a 3,4,5 triangle
• That is., hypotenuse of ⊿ADC is 5 cm
5. The centre of the circle bisects the diagonal. 
So radius of the circle = ^AC⁄₂ = ⁵⁄₂ = 2.5
6. So area of the circle = πr² = π × 2.5² = 6.25π.

Solved example 27.7
How do we draw a triangle with two of the angles 40^o and 120^o within a circle of 3 cm radius?
Solution:
1. First step is to draw a rough sketch. This is shown in fig.27.24(a) below.
• Let A, B and C be the three vertices. All of them lie on a circle
• The third angle can be calculated as: [180 - (120+40)] = [180 -120-40] = 20^o. But we don't really need it

Fig.27.24

2. Join A and B to the centre O
3. Separate out the minor arc AB. This is shown in fig(b)
4. Now we have an arc AB which subtends ∠ACB on the alternate arc.
5. ∠ACB = 120^o. So, using the converse of theorem 27.4, we get: ∠AOB = 2 × 120 = 240. This is marked in fig(b)
6. Again separate out the minor arc BC. This is shown in fig(c)
7. Now we have an arc BC which subtends ∠BAC on the alternate arc.
8. ∠BAC = 40^o. So, using the converse of theorem 27.4, we get: ∠BOC = 2 × 40 = 80. This is marked in fig(c)
9. So now we have the angles between the radial lines from centre O. We can do the actual construction. 
10. At any convenient point O, draw a circle of radius 3 cm
11. Draw a radial line OA in any convenient direction. Draw OB such that ∠OAB is 240^o. This is shown in fig.27.25 below:

Fig.27.25

11. Next draw OC such that ∠BOC is 80^o.
12. Thus the three vertices are obtained. Draw AB, BC and CA to get the required ΔABC.

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In the next section, we will see a few more solved examples.

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Chapter 27.2 - Central angle and Subtended angle of Arcs

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:

Fig.27.14

1. Fig(a) shows the major arc AXB. It has a central angle c^o
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 x^o
• Then ∠PAO = ∠APO = x^o. These are shown in cyan colour
• ∠AOP = (180 -2x)^o  (∵ sum of interior angles of a triangle is 180^o) 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 y^o
• Then ∠PBO = ∠BPO = y^o. These are shown in white colour
• ∠ BOP = (180 -2y)^o  (∵ sum of interior angles of a triangle is 180^o) 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 = c^o. 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 360^o. 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 c^o
(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

In the fig.27.15(a) above:
(i) We are given an arc AXB. It has a central angle of 240^o
(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 = 120^o.
■ Figs. (b) and (c) shows that P can be any where on the arc AXB. The angle 120^o will not change.  
11. The above fact can be used to calculate the angle at P. Let us see some examples:

Fig.27.16

In fig.27.16(a) above:
(i) We are given an arc AB (magenta coloured). It has a central angle of 160^o
(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 = 80^o 
In fig.27.16(b) above:
(i) We are given an arc AB (magenta coloured). It has a central angle of 108^o
(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 = 54^o 
In fig.27.16(c) above:
(i) We are given an arc AB (magenta coloured). It has a central angle of 70^o
(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 = 35^o.

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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 c^o.
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 a^o.
4. Then the central angle c^o of the original arc is given by: c^o = 2 × a^o

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A very interesting case:

1. Consider fig.27.17 below:

Fig.27.17

2. In fig(a), an arc AYB has a central angle of c^o. 
• 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

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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 90^o.

Fig.27.18

2. Separate out an arc ABC. This is shown in fig(b).
3. This arc ABC subtends an ∠ADC = 90^o, 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 × 90^o) = 180^o.  
5. If ∠AOC is 180^o, 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.

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The above findings can be used as a standard result. Let us write the steps:

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. 

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In the next section, we will see a few more solved examples.

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