Chapter 27.11 - Rectangle into Square of Equal area

In the previous section we saw an application of theorem 27.10 . We also saw some solved examples. In this section we will see a special case of that application.

■ In the previous section we saw this:
• A rectangle was given to us
    ♦ We made a new rectangle of the same area but different dimensions
■ In this section we will see this:
• A rectangle will be given to us
    ♦ We will make a square of the same area.
■ We will learn the method by analysing an actual example:

Solved example 27.22

Draw a rectangle of Length 4 cm and width 2 cm. Draw a square of the same area.

Solution:

Consider the rectangle in fig.27.59(a) below:

Fig.27.59

• It has a length of 4 cm and width of 2 cm. We want to change it into a square. 
• But there is one condition: The area of the rectangle and the new square must be the same. 
• Let the side of the new square be 'x'. It is shown in fig.27.59(b). We want to find this 'x' graphically. Let us try:
• Consider fig27.59(c). Two chords AB and CD intersect at P. 
• Out of the two chords, AB is a diameter. Because it is passing through the centre 'O'
• This diameter intersects the other chord CD in a perpendicular direction. So CD is bisected. 
    ♦ That means, PC = PD = x. See theorem 17.1.
• The lengths of the four pieces are:
PA = 2, PB = 4, PC = x, PD = x
• Imagine a rectangle with length 4 and width 2
    ♦ It's area will be equal to 4×2 = PA×PB
• Imagine another rectangle with length x and width x  
    ♦ It's area will be equal to x×x = PD×PC
■ Based on theorem 27.10 , the two areas will be equal.
• A rectangle with length x and width also x is a square.
• So our next aim is to construct a circle and the chords shown in fig.27.59(c). 
• What we have, is the given rectangle with length 4 cm and width 2 cm. We have to begin our work from that rectangle.

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1. Consider fig.27.60(a) below. The base of the given rectangle is named as PB. So PB = 4. For the ease of construction, the given rectangle should be placed in such a way that the side PB is exactly horizontal.

Fig.27.60

2. With P as centre and PA as radius, draw an arc (shown in yellow colour) which will cut the horizontal through P at A. So PA = 2 cm

3. With AB as diameter, draw a circle. This is shown in fig(b).
• For drawing the circle, first draw the perpendicular bisector of AB. It will cut AB at centre 'O'. 
• With centre 'O' and OA as radius, draw the circle. This step is not shown in the fig.
4. Draw a vertical through P. It will intersect the circle at C and D.
• AB is a diameter of the circle. CD is a chord
• Since PB is horizontal and CD is vertical, ∠PBD = 90^o
• So diameter AB is perpendicular to CD. Then by theorem 17.1, AB is the perpendicular bisector of CD. Thus we get PC = PD
5. Once we get any one side of a square, we can easily construct it. Here, Both PC and PD are equal to the side of the required square. We can use any one of them. 
• We will use the lower PD So that the square will be distinct from the given rectangle
• Thus in the fig(c), the square PP'ED is constructed

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Now, there is an easier method to obtain the required square:

Consider fig.27.60(c) above. Our real aim is to obtain the side PD. If we can obtain this side directly, a lot of work can be saved. Let us see the method for doing it:

Consider fig.27.61(a) below:

Fig.27.61

1. Draw a line AB, 6 cm in length

2. Mark a point P such that PA = 2 cm and PB = 4 cm

3. Draw a semi-circle with AB as diameter

• For drawing the circle, first draw the perpendicular bisector of AB. It will cut AB at centre 'O'. 
• With centre 'O' and OA as radius, draw the circle. This step is not shown in the fig.
4. Draw a perpendicular to AB through P. This perpendicular will meet the semi-circle at D
5. PD is the required side. The square PP'ED can then be easily drawn.
■ In this method, we do not even have to draw the original rectangle
■ Also note that, the semi-circle can be drawn on the upper side of AB 

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Solved example 27.23

Let a rectangle be of length 6 cm and width 4 cm. Draw a square of the same area.

Solution:

1. Draw a line AB, (6+4) = 10 cm in length. See fig.27.61(b) above.

2. Mark a point P such that PA = 4 cm and PB = 6 cm

3. Draw a semi-circle with AB as diameter

4. Draw a perpendicular to AB through P. This perpendicular will meet the semi-circle at D
5. PD is the required side. The square PP'ED can then be easily drawn.

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Irrational numbers

Let us analyse the above two solved examples.
(i) In the solved example 27.22, we got a square whose area is same as a rectangle of length 4 cm and width 2 cm
• So the area of the newly formed square is 8 cm².
• If the area of a square is 8 cm², then obviously, it's side would be √8 cm
• Thus the length of PD = √8 cm 
(ii) In the solved example 27.23, we got a square whose area is same as a rectangle of length 6 cm and width 4 cm
• So the area of the newly formed square is 24 cm².
• If the area of a square is 24 cm², then obviously, it's side would be √24 cm 
• Thus the length of PD = √24 cm 
■ So this is an excellent method to find the values of irrational numbers graphically. 
• Note that in an earlier chapter, we had learned another method for doing this. Details here.

Another example:
Consider fig.27.62(a) below:

Fig.27.62

1. Out of the two chords, chord AB is a diameter. Chord CD is drawn perpendicular to AB. 
• So AB will be the perpendicular bisector of CD. Thus PC = PD
2. Now let us apply theorem 27.10:
Multiplying opposite pieces of the same chord, we get: 
PA×PB = PC×PD ⇒ PA×PB = PC² (∵ PC = PD)
3. This is a very useful result. 
• We no longer need the piece PD. So we need not consider the lower part of the circle. 
• That means, all our further calculations will be related to the portion above the diameter AB. 
• That is., we will be dealing with a semi-circle only.
4. In fig.27.62(b) above, the diameter AB (= 8 cm) is split into two parts at 'P'. 
PA is 6 cm and PB is 2 cm. 
5. A perpendicular PC is erected at P in such a way that C lies on the semi-circle with AB as diameter. 
6. Using the equation in (2) above, we get:
 PA×PB = PC². That is., 6×2 = PC² ⇒ 12 = PC² ⇒ PC = √12
■ Let us write a summary of what we have done above:
• We drew a semi-circle with diameter AB = 8 cm
• We split the diameter at P such that PA = 6 cm and PB = 2 cm
• Finally we erected a perpendicular at P in such a way that it meets the circle at C
• We find that PC = √12 cm

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• We know that √12 is an irrational number. We cannot obtain √12 on a scale. 
• But using the above method, we are able to draw a line of length √12 cm with out using a scale.    
• Once we obtain a line of length √12 cm, if required, we can construct a square of area 12 cm². 
• Because area of a square of side √12 cm = √12 ×√12 = 12 cm²
■ The procedure for constructing the square is shown in fig(c)
1. First draw a horizontal line through C
2. Then draw an arc with C as centre and CP as radius. This arc will cut the horizontal line at a point. Name this point as D
3. Through D, drop a perpendicular to AB. Name the foot of this perpendicular as F
4. Then FPCD is a square of area 12 cm²

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Now let us see some solved examples: 

Solved example 27.24

Find the value of 'x' in the figs.27.63(a), (b) and(c) below:

Fig.27.63

Solution:
Case (a): x² = 8×18 ⇒ x² = 144 ⇒ x = √144 ⇒ x = 12
Case (b): x² = 9×4 ⇒ x² = 36 ⇒ x = √36 ⇒ x = 6
Case (c): x² = 2×5 ⇒ x² = 10 ⇒ x = √10

Solved example 27.25

In the fig.27.63(d) above, AD = 10 cm, BD = 6 cm, CD = 2 cm. Find the value of CP, CQ and PQ.

Solution:
• CQ² = AC×CD = (AD-CD)×CD = (10-2)×2 = 8×2 = 16 cm
⇒ CQ = √16 = 4 cm 
• CP² = BC×CD = (BD-CD)×CD = (6-2)×2 = 4×2 = 8 cm
⇒ CP = √8 = 2√2 cm 
• PQ = (CQ - CP) = (4 - 2√2) = 2(2-√2) cm

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A video presentation of a problem can be seen here:

Trapezium problem

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

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Chapter 27.10 - Rectangles of Equal areas

In the previous section we saw theorem 27.10 and it's application. We also saw some solved examples. In this section we will see another application of the theorem.

Consider the two chords AB and CD in fig.27.55(a) below. They intersect at P.

Fig.27.55

1. Consider the chord AB. It is split into two pieces PA and PB. 
2. Let us construct a rectangle with sides PA and PB. The longer piece PA will be the length and shorter piece PB will be the width. (In fact, the results will not be affected even if lengths and widths are interchanged) 
3. Let us construct the rectangle over the circle itself. So the longer side of the rectangle will coincide with PA. This is shown in fig(b). 
4. The width of the rectangle must be exactly equal to PB. For that, with P as centre and PB as radius, draw an arc (shown in yellow colour). Let this arc intersect the perpendicular from P at P'. Then PP' is the width of our rectangle. So The rectangle APP'A' can be easily completed.
5. Area of rectangle APP'A' = AP×PP' = PA×PB

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6. Next, consider the chord CD. It is split into two pieces PC and PD. 

7. Let us construct a rectangle with sides PC and PD. The longer piece PD will be the length and shorter piece PC will be the width. (In fact, the results will not be affected even if lengths and widths are interchanged) 
8. Let us construct the rectangle over the circle itself. So the shorter side of the rectangle will coincide with PC. This is shown in fig(c). 
9. The width of the rectangle must be exactly equal to PD. For that, with P as centre and PD as radius, draw an arc (shown in blue colour). Let this arc intersect the perpendicular from P at P''. Then PP'' is the length of our rectangle. So The rectangle CPP''C' can be easily completed.
10. Area of rectangle CPP''C' = CP×P''P = PC×PD
■ But applying theorem 27.10 , PA×PB = PC×PD. So the results in (5) and (10) are equal. Thus we find that, the areas of the two rectangles are equal.

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Now let us put the above result for a practical application:
Consider the rectangle in fig.27.56(a) below. 

Fig.27.56

• It has a length of 'a' and width of 'b'. We want to increase it's length by 'c'. So the new length would be '(a+c)'. 
• But there is one condition: The area of the rectangle must remain the same. 
• If the area is to remain the same, obviously, the width will have to decrease. Let the new width be 'x'. It is shown in fig.27.56(b). We want to find this 'x' graphically. Let us try:
• Consider fig27.56(c). Two chords AB and CD intersect at P. The lengths of the four pieces are:
PA = b, PB = a, PC = x, PD = (a+c)
• Imagine a rectangle with length a and width b
    ♦ It's area will be equal to ab = PA×PB
• Imagine another rectangle with length (a+c) and width x  
    ♦ It's area will be equal to (a+c)x = PD×PC
■ Based on theorem 27.10 , the two areas will be equal.
• So our next aim is to construct a circle and the chords shown in fig.27.56(c). 
• What we have is the given rectangle with length a and width b. We have to begin our work from that rectangle.

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1. Consider fig.27.57(a) below. The base of the given rectangle is named as PB. So PB = a. For the ease of construction, the given rectangle should be placed in such a way that the side PB is exactly horizontal.

Fig.27.57

2. Next, PB is extended towards the right by a distance 'c', thus reaching the point B'. So PB' = (a+c)

3. With P as centre and PB' as radius, draw an arc (shown in red colour). It will intersect the vertical through P at D. So PD = (a+c). 

• Note that, PB is horizontal and PD is vertical. So ∠BPD = 90^o.

4. With P as centre and b as radius, draw an arc (shown in yellow colour). It will intersect the horizontal through P at A. So PA = b
5. Thus we get the three points, A, B and D. Note that these are the same A, B and D in the fig.27.56(c) that we saw earlier. 
6. In that fig., a circle passes through those three points. 
• A circle passing through any three points is unique. That means there is one and only one circle which will pass through three points. 
• We have learned about it earlier. See details here. Also we know how to draw the circle passing through any given three points. 
7. So, once we have the three points A, B and D, we can draw the circle through them. This is shown in fig.27.57(b). 
8. The circle will cut the vertical through P at C. PC will naturally be equal to 'x'. So we have drawn the circle and the chords in the earlier fig. 27.56(c). Now we can proceed to draw the new rectangle. 
9. With P as centre and PC as radius, draw an arc (shown in blue colour in fig.27.57.c). It will intersect the horizontal through P at P'. So PP' = x.
10. So PP' is the width of the required rectangle. The length (=PD) is already available. Thus we can easily construct the new rectangle  EDPP'

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Solved example 27.20

Draw a rectangle of Length 6 cm and width 4 cm. Draw a rectangle of the same area with length 7 cm.

Solution:

The required construction is shown in fig.27.58(a) below:

Fig.27.58

Let us see the steps:

1. Draw a rectangle of length 6 cm and width 4 cm. The length is named as PB. For the ease of construction, PB must be perfectly horizontal.
2. Extend PB towards the right by 1 cm upto B'. So PB' = 7 cm
3. With P as centre and PB' as radius, draw an arc (shown in red colour). It will intersect the vertical through P at D. So PD = 7 cm.

• Note that, PB is horizontal and PD is vertical. So ∠BPD = 90^o.

4. With P as centre and 4 cm as radius, draw an arc (shown in yellow colour). It will intersect the horizontal through P at A. So PA = 4 cm 
5. Thus we get the three points, A, B and D.
6. Once we have the three points A, B and D, we can draw the circle through them.
7. The circle will cut the vertical through P at C. PC will naturally be equal to 'x'. 
8. With P as centre and PC as radius, draw an arc (shown in blue colour). It will intersect the horizontal through P at P'. So PP' = x.
9. So PP' is the width of the required rectangle. The length (=PD) is already available. Thus we can easily construct the new rectangle  EDPP'

Solved example 27.21
Draw a rectangle of Length 7 cm and width 3 cm. Draw a rectangle of the same area with length 5 cm.

Solution:

The required construction is shown in fig.27.58(b) above.

Let us see the steps:

1. Draw a rectangle of length 7 cm and width 3 cm. The length is named as PB. For the ease of construction, PB must be perfectly horizontal.
2. In this problem, length of the new rectangle is less. So there is no need for extension. We can mark B' within PB. We mark B' in such a way that PB' = 5 cm
3. With P as centre and PB' as radius, draw an arc (shown in red colour). It will intersect the vertical through P at D. So PD = 5 cm.

• Note that, PB is horizontal and PD is vertical. So ∠BPD = 90^o.

4. With P as centre and 3 cm as radius, draw an arc (shown in yellow colour). It will intersect the horizontal through P at A. So PA = 4 cm 
5. Thus we get the three points, A, B and D.
6. Once we have the three points A, B and D, we can draw the circle through them.
7. The circle will cut the vertical through P at C. PC will naturally be equal to 'x'. 
8. With P as centre and PC as radius, draw an arc (shown in blue colour). It will intersect the horizontal through P at P'. So PP' = x.
9. So PP' is the width of the required rectangle. The length (=PD) is already available. Thus we can easily construct the new rectangle  EDPP'

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So we saw an application of theorem 27.10. In the next section, we will see a special case of this application.

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Chapter 27.9 - Chords inside a Circle

In the previous section we completed the discussion on cyclic quadrilaterals. In this section we will see chords. We saw some basic details about chords in chapter 17.3. Now we will see some advanced details.

• Consider any two chords of a circle. Only condition is that, they must be non-parallel. 
• Since they are non-parallel, they will surely intersect at a point 'P'. 
• This 'P' may be inside the circle as shown in fig.27.49(a). Or 'P' may be outside the circle as shown in fig.27.49(b).

Fig.27.49

■ Which ever be the case, there are some similarities between the two. Let us analyse:
1. Consider fig.27.50(a) below. It is the same fig.27.49(a) that we saw above. 

Fig.27.50

2. A small modification is made. That is., AD and BC are drawn with red lines. That is., ends  of one chord are joined to the ends of the other. 
3. Now we get two triangles: ΔAPD and ΔBPC. 
• These two triangles are similar. We can prove this as follows:
4. In the fig.27.51(a) below, a chord BD is drawn. 

Fig.27.51

5. This chord BD divides the circle into two segments. Take out the larger segment. 
6. ∠DAB and ∠DCB are two angles in this larger segment. So they are both equal to the unique angle. See theorem 27.7. That means both have the same angle value. 
7. So we can write this:    
• ∠DAP in ΔAPD is equal to ∠BCP in ΔBPC
• In other word, ∠A in ΔAPD is equal to ∠C in ΔBPC (They are shown in yellow colour)
8. Let us see if there are any other equal angles like them:
• Consider ∠APD and ∠BPC. They are opposite angles, and hence equal. So we can write:
• ∠P in ΔAPD is equal to ∠P in ΔBPC (They are shown in white colour)
9. Thus we get 'two angles the same' in the two triangles ΔAPD and ΔBPC. 
• Naturally the third angle must also be the same. 
However, we will write the calculation steps:
(i) The third angle (ie., ∠D) in ΔAPD = 180 - ∠A - ∠P 
(ii) The third angle (ie., ∠B) in ΔBPC = 180 - ∠C - ∠P 
(iii) But ∠C = ∠A and ∠P = ∠P. So (i) will be equal to (ii). That is., ∠D will be equal to ∠B
10. Thus all the angles in the two triangles ΔAPD and ΔBPC are equal
■ So they are similar triangles
11. Now we apply a special property that is applicable to any two similar triangles:
^{side opposite smallest angle in ΔAPD}⁄_{side opposite smallest angle in ΔBPC} 
= ^{side opposite medium angle in ΔAPD}⁄_{side opposite medium angle in ΔBPC} 
= ^{side opposite largest angle in ΔAPD}⁄_{side opposite largest angle in ΔBPC}
12. But angles in the two triangles are the same. That is.,
• Smallest angle in ΔAPD = Smallest angle in ΔBPC
• Medium angle in ΔAPD = Medium angle in ΔBPC
• Largest angle in ΔAPD = Largest angle in ΔBPC
13. So we can write this:
• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,
^{side opposite ∠A in ΔAPD}⁄_{side opposite ∠C in ΔBPC} 
= ^{side opposite ∠D in ΔAPD}⁄_{side opposite ∠B in ΔBPC}
= ^{side opposite ∠P in ΔAPD}⁄_{side opposite ∠P in ΔBPC}
14. So we get: ^PD⁄_PB = ^AP⁄_PC = ^AD⁄_BC .

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Now let us see if we can derive the same result in (14) for the case when P is outside the circle:

1. Consider fig.27.50(b) above. It is the same fig.27.49(b) that we saw earlier. 
2. A small modification is made. That is., AD and BC are drawn with red lines. That is., ends  of one chord are joined to the ends of the other. 
3. Now we get two triangles: ΔAPD and ΔBPC. [Note that, even though 'P' is outside the circle, we get triangles with the same names as in the previous case] 
• These two triangles are similar. We can prove this as follows:
4. In the fig.27.51(b) above, a chord BD is drawn.
5. This chord BD divides the circle into two segments. Take out the larger segment. 
6. ∠DAB and ∠DCB are two angles in this larger segment. So they are both equal to the unique angle. See theorem 27.7. That means both have the same angle value. 
7. So we can write this:    
• ∠DAP in ΔAPD is equal to ∠BCP in ΔBPC
• In other word, ∠A in ΔAPD is equal to ∠C in ΔBPC (They are shown in yellow colour)
8. Let us see if there are any other equal angles like them:
• Consider ∠APD and ∠BPC. They are one and the same, and hence equal. So we can write:
• ∠P in ΔAPD is equal to ∠P in ΔBPC (This is shown in white colour)
9. Thus we get 'two angles the same' in the two triangles ΔAPD and ΔBPC. 
• Naturally the third angle must also be the same. 
However, we will write the calculation steps:
(i) The third angle (ie., ∠D) in ΔAPD = 180 - ∠A - ∠P 
(ii) The third angle (ie., ∠B) in ΔBPC = 180 - ∠C - ∠P 
(iii) But ∠C = ∠A and ∠P = ∠P. So (i) will be equal to (ii). That is., ∠D will be equal to ∠B
10. Thus all the angles in the two triangles ΔAPD and ΔBPC are equal
■ So they are similar triangles
11. Now we apply a special property that is applicable to any two similar triangles:
^{side opposite smallest angle in ΔAPD}⁄_{side opposite smallest angle in ΔBPC} 
= ^{side opposite medium angle in ΔAPD}⁄_{side opposite medium angle in ΔBPC} 
= ^{side opposite largest angle in ΔAPD}⁄_{side opposite largest angle in ΔBPC}
12. But angles in the two triangles are the same. That is.,
• Smallest angle in ΔAPD = Smallest angle in ΔBPC
• Medium angle in ΔAPD = Medium angle in ΔBPC
• Largest angle in ΔAPD = Largest angle in ΔBPC
13. So we can write this:
• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,
^{side opposite ∠A in ΔAPD}⁄_{side opposite ∠C in ΔBPC} 
= ^{side opposite ∠D in ΔAPD}⁄_{side opposite ∠B in ΔBPC}
= ^{side opposite ∠P in ΔAPD}⁄_{side opposite ∠P in ΔBPC}
14. So we get: ^PD⁄_PB = ^AP⁄_PC = ^AD⁄_BC.

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So we get the same result (14) in both the cases. That means, the result is valid for both 'P inside' and 'P outside' the circle.
1. Now, from among the three ratios, we will take out two, which has 'P'. So we take out the first and second. So we get:
^PD⁄_PB = ^AP⁄_PC  
2. Cross multiplying we get: PA × PB = PC × PD

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So we can always multiply the opposite pieces. We will write this result as a theorem.

Theorem 27.10:

1. Two chords of a circle meet at a point inside the circle

2. The point divides the each chord into two pieces

3. Multiply the two pieces belonging to one chord

4. Multiply the two pieces belonging to the other chord

5. The two products will always be equal

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Let us now see one application of the above theorem. We will see it as a solved example:

Solved example 27.18
The distance between the ends of a piece of bangle is 4 cm. It’s height is 1 cm. What is the radius of the full bangle?
Solution:
We did this problem when we learned about length of chords. See here. Now we will do it using another method:
1. In the fig.27.53(a) below, A and B are the ends of the bangle. The distance between them is 4 cm.

Fig.27.53

2. The height of the piece is given as 1 cm. It should be measured in a direction perpendicular to the line AB. This is also shown in fig.27.53(a)
3. In fig(b), the remaining portion of the bangle is shown in dashed line.  
4. AB is a chord of the full circle. 
5. Consider a diameter CD. It must satisfy one condition:
It must be perpendicular to the chord AB
6. If this condition is satisfied, the diameter CD will bisect the chord AB at P. See Theorem 17.1.
7. When AB is bisected, AP = BP = 2 cm
8. Since the diameter is also a chord, we can apply theorem 27.10. Thus,
Multiplying opposite pieces of the same chord, we get:
PA×PB = PD×PC ⇒ 2×2 = 1×PC ⇒ PC = 4
9. So diameter of the bangle = CD = PD + PC = 1 + 4 = 5 cm
10. So radius of the bangle = ⁵⁄₂ = 2.5 cm

Solved example 27.19
Find the value of 'x' in each of the three cases in fig.27.54 below:

Fig.27.54

Solution:
1. Consider fig(a). We can multiply opposite pieces: 16×6 = x×12 ⇒ 96 = 12x ⇒ x = 8 
2. Consider fig(b). We can multiply opposite pieces: 20×6 = x×8 ⇒ 120 = 8x ⇒ x = 15
3. Consider fig(c). We can multiply opposite pieces: But length of one piece is not given
■ We are given two clues:
• The chord with a total length of 13 cm, is a diameter. Because it is passing through the centre 'O'
• This diameter intersects the other chord in a perpendicular direction. So the other chord is bisected. 
    ♦ That means, the length of the other piece is also 'x'. See Theorem 17.1.
So we can write: 9×4 = x×x ⇒ 36 = x² ⇒ x = 6

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In the next section, we will see another application of theorem 27.10.

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Chapter 27.8 - Cyclic quadrilaterals - Solved examples

In the previous section we saw theorem 27.9 and it's converse. In this section we will see some solved examples.

Solved example 27.15
In the fig.27.46(a) below, A, B, C and D are four points on a circle. 

Fig.27.45

∠DBC = 55^o and ∠CAB = 45^o. Compute ∠BCD
Solution:

1. Consider arc BC in fig(b). It subtends ∠BAC (= 45^o) on the alternate arc. 

• The same arc subtends ∠BDC on the alternate arc. So ∠BDC = ∠BAC = 45^o. This is marked in fig(b)
2. Consider ΔBDC. We get ∠BCD = [180-(45+55)] =[180-100] = 80^o. (∵ sum of interior angles of a triangle is 180^o)
Thus we get the required angle. We can do a check by using theorem 27.9.   

3. Consider arc CD in fig(c). It subtends ∠CBD (= 55^o) on the alternate arc. 

• The same arc subtends ∠CAD on the alternate arc. So ∠CBD = ∠CAD = 55^o. This is marked in fig(c)
(i) Sum of opposite angles ∠BAD and ∠BCD = 45 + 55 + 80 = 180^o.          

Solved examples 27.16
In the fig.27.46(a) below, A, B, C and D are four points on a circle. 

Fig.27.46

AC and BD intersect at E in such a way that ∠BEC = 30^o. and ∠ECD = 20^o. Find ∠BAC
Solution:
1. ∠BEC and ∠BEA form a linear pair. So ∠BEA = 180 - ∠BEC = 180 -130 = 50^o.
2. ∠BEA and ∠CED are opposite angles, and are hence equal. So we get ∠BEA = ∠CED = 50^o.

3. Consider ΔCED. We get ∠EDC = [180-(50+20)] =[180-70] = 110^o. (∵ sum of interior angles of a triangle is 180^o)

4. Consider the major arc BC in fig(c). It subtends ∠BDC (= 110^o) on the alternate arc. 

• The same arc subtends ∠BAC on the alternate arc. So ∠BAC = ∠BDC = 110^o. This is marked in fig(c). Thus we get the required angle.

Solved example 27.17
In the fig.27.47(a) below, ABCD is a square. 

Fig.27.47

Determine ∠DPC
Solution: 
1. Draw the diagonal AC of the square. 
2. A diagonal of a square will bisect the angles at the corners. So we get:
∠DAC = ∠BAC = 45^o.
3. Consider the quadrilateral ACPD. It is a cyclic quadrilateral. The sum of opposite angles = 180^o.
4. So we get: ∠DPC + ∠DAC = 180^o ⇒ ∠DPC + 45 = 180 ⇒ ∠DPC = 180 - 45 = 135^o.

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Now we will see an important result related to cyclic quadrilaterals. We will learn it in steps: 

1. Fig.27.48(a) below shows a cyclic quadrilateral ABCD. 

Fig.27.48

2. The side AB is extended along towards the right up to point E. So ∠CBE (shown in red colour) becomes an exterior angle of the cyclic quadrilateral. We can write:
■ ∠CBE is the exterior angle of the cyclic quadrilateral ABCD at the vertex B. 
3. For the vertex B, the opposite vertex is D
• So, for the vertex B, ∠ADC (shown in yellow colour) is the 'interior angle at the opposite vertex'
4. Thus we have three quantities:
(i) A vertex B  (ii) Exterior angle at that vertex B  (iii) Interior angle at D, which is the opposite vertex of B  
• We want to know the relation between (ii) and (iii)
5. Consider the interior angle at B. It is shown in blue colour in fig (b)
• Blue + Red will obviously be 180^o (∵ they form a linear pair)
• So we can write: ∠CBE + ∠ABC = 180^o
6. Yellow and Blue are opposite angles of a cyclic quadrilateral. So their sum will be 180^o. 
• We can write: ∠ADC + ∠ABC = 180^o
7. From (5) we get: ∠ABC = 180 – ∠CBE
• Substituting this in (6) we get:
∠ADC + (180 – ∠CBE) = 180
⇒ ∠ADC – ∠CBE = 180 – 180  
⇒ ∠ADC – ∠CBE = 0 
⇒ ∠ADC = ∠CBE
8. So we can write: 
• The exterior angle at B is equal to the interior angle at opposite vertex
9. Now consider fig (c). The side CB is extended upto F
• ∠ABF is an exterior angle at vertex B. So is ∠CBE
• But we can see that the two are equal because, they are opposite angles. 
• So, at a vertex, there will be only one value for an exterior angle
10. We can write the above results in a general form:
• Consider any vertex of a cyclic quadrilateral. 
• There will be an exterior angle at that vertex
• That exterior angle will be equal to the interior angle at the opposite vertex

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Some solved examples on cyclic quadrilaterals are shown in the form of a video presentation at the following links:
Trapezium Cyclic or not

Non-rectangular parallelogram Cyclic or not

Video presentation 1 

Video presentation 2 

Video presentation 3 

Video presentation 4 

Parallelogram with one vertex at center

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In the next section, we will learn about Multiplication of Chords.

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