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).
■ 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.
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.
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 .
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.
• 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 |
1. Consider fig.27.50(a) below. It is the same fig.27.49(a) that we saw above.
 |
| Fig.27.50 |
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 |
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 .
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.
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
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.
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:
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
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
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
Let us now see one application of the above theorem. We will see it as a solved example:
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 |
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:
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| Fig.27.54 |
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
In the next section, we will see another application of theorem 27.10.
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