In the previous section we completed the discussion on chords which intersect at a point inside the circle. In this section we will see chords which intersect at a point outside the circle. We have earlier seen that even if the point of intersection is outside the circle, the segments can be multiplied (Details here). Now we will see an easier derivation:
Consider fig.27.68(a) below.
• AB and CD are two chords of a circle. If they are extended towards the right, they will meet at a point P outside the circle.
• From the point P, we can consider four segments. They are: PA, PB, PC and PD.
• We want to know the relations between these four segments. Let us analyse:
1. In the fig (b), AC and BD are drawn. We can consider two triangles: △PAC and △PBD
2. In both the triangles, the angle at P is the same. It is shown in blue colour.
3. Now consider the cyclic quadrilateral ABDC. It has an exterior angle at vertex B. It is ∠DBP
4. But this exterior angle will be equal to the interior angle at opposite vertex C (Details here)
5. So we have ∠DCA = ∠DBP. Both are shown in red colour
6. In △PAC, the third angle ∠PAC = 180 – (red + blue)
7. In △PBD, the third angle ∠PBD = 180 – (red + blue)
8. So it is clear that ∠PAC = ∠PBD. They are shown in cyan colour
9. We have two triangles △PAC and △PBD with the same angles.
■ So they are similar triangles
10. Now we apply a special property that is applicable to any two similar triangles (Details here):
side opposite smallest angle in ΔPAC⁄side opposite smallest angle in ΔPBD
= side opposite medium angle in ΔPAC⁄side opposite smallest angle in ΔPBD
= side opposite largest angle in ΔPAC⁄side opposite smallest angle in ΔPBD
11. But angles in the two triangles are the same. That is.,
• Smallest angle in ΔPAC = Smallest angle in ΔPBD
• Medium angle in ΔPAC = Medium angle in ΔPBD
• Largest angle in ΔPAC = Largest angle in ΔPBD
12. So we can write this:
• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,
side opposite red angle in ΔPAC⁄side opposite red angle in ΔPBD
= side opposite blue angle in ΔPAC⁄side opposite blue angle in ΔPBD
= side opposite cyan angle in ΔPAC⁄side opposite cyan angle in ΔPBD
13. So we get: PA⁄PD = AC⁄BD = PC⁄PB .
14. Let us take out the two ratios which has 'P' in them:
PA⁄PD = PC⁄PB .
15. Cross multiplying we get: PA × PB = PC × PD
• This is a very useful result. It is easy to remember in this way:
1. Two lines are drawn from an exterior point P towards a circle
• One of them cuts the circle at two points A and B
• The other cuts the circle at two points C and D
2. Consider the line which cuts the circle at A and B
(i) It's whole length is PA
(ii) It's length outside the circle is PB
• Take the product. That is:
(Whole length) × (Length outside) = PA×PB
3. Consider the line which cuts the circle at C and D
(i) It's whole length is PC
(ii) It's length outside the circle is PD
• Take the product. That is:
(Whole length) × (Length outside) = PC×PD
■ The products in (2) and (3) will be equal
Consider fig.27.68(a) below.
• AB and CD are two chords of a circle. If they are extended towards the right, they will meet at a point P outside the circle.
 |
| Fig.27.68 |
• We want to know the relations between these four segments. Let us analyse:
1. In the fig (b), AC and BD are drawn. We can consider two triangles: △PAC and △PBD
2. In both the triangles, the angle at P is the same. It is shown in blue colour.
3. Now consider the cyclic quadrilateral ABDC. It has an exterior angle at vertex B. It is ∠DBP
4. But this exterior angle will be equal to the interior angle at opposite vertex C (Details here)
5. So we have ∠DCA = ∠DBP. Both are shown in red colour
6. In △PAC, the third angle ∠PAC = 180 – (red + blue)
7. In △PBD, the third angle ∠PBD = 180 – (red + blue)
8. So it is clear that ∠PAC = ∠PBD. They are shown in cyan colour
9. We have two triangles △PAC and △PBD with the same angles.
■ So they are similar triangles
10. Now we apply a special property that is applicable to any two similar triangles (Details here):
side opposite smallest angle in ΔPAC⁄side opposite smallest angle in ΔPBD
= side opposite medium angle in ΔPAC⁄side opposite smallest angle in ΔPBD
= side opposite largest angle in ΔPAC⁄side opposite smallest angle in ΔPBD
11. But angles in the two triangles are the same. That is.,
• Smallest angle in ΔPAC = Smallest angle in ΔPBD
• Medium angle in ΔPAC = Medium angle in ΔPBD
• Largest angle in ΔPAC = Largest angle in ΔPBD
12. So we can write this:
• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,
side opposite red angle in ΔPAC⁄side opposite red angle in ΔPBD
= side opposite blue angle in ΔPAC⁄side opposite blue angle in ΔPBD
= side opposite cyan angle in ΔPAC⁄side opposite cyan angle in ΔPBD
13. So we get: PA⁄PD = AC⁄BD = PC⁄PB .
14. Let us take out the two ratios which has 'P' in them:
PA⁄PD = PC⁄PB .
15. Cross multiplying we get: PA × PB = PC × PD
• This is a very useful result. It is easy to remember in this way:
1. Two lines are drawn from an exterior point P towards a circle
• One of them cuts the circle at two points A and B
• The other cuts the circle at two points C and D
2. Consider the line which cuts the circle at A and B
(i) It's whole length is PA
(ii) It's length outside the circle is PB
• Take the product. That is:
(Whole length) × (Length outside) = PA×PB
3. Consider the line which cuts the circle at C and D
(i) It's whole length is PC
(ii) It's length outside the circle is PD
• Take the product. That is:
(Whole length) × (Length outside) = PC×PD
■ The products in (2) and (3) will be equal
Now we will see an interesting result:
1. In fig.27.68(a) above, we have: PA × PB = PC × PD
2. What if PB = PD?
3. If PB = PD, we will get PA = PC
4. AB = PA - PB ⇒ AB = PC - PD (∵ PA = PC & PB = PD)
5. CD = PC - PD
6. From (4) and (5) we get AB = CD
7. Now consider the △PAC. In (3) we got PA = PC
8. So △PAC is an isosceles triangle. Then it's base angles will be equal. That is., ∠DCA = ∠PAC
9. Now consider the quadrilateral ABDC
• From (6) we have AB = CD
• From (8) we have ∠DCA = ∠PAC
• So ABDC is an isosceles trapezium
1. In fig.27.68(a) above, we have: PA × PB = PC × PD
2. What if PB = PD?
3. If PB = PD, we will get PA = PC
4. AB = PA - PB ⇒ AB = PC - PD (∵ PA = PC & PB = PD)
5. CD = PC - PD
6. From (4) and (5) we get AB = CD
7. Now consider the △PAC. In (3) we got PA = PC
8. So △PAC is an isosceles triangle. Then it's base angles will be equal. That is., ∠DCA = ∠PAC
9. Now consider the quadrilateral ABDC
• From (6) we have AB = CD
• From (8) we have ∠DCA = ∠PAC
• So ABDC is an isosceles trapezium
Now we will see some solved examples
Solved example 27.29
Find the value of x in each of the figs given below:
Solution:
(a) We have: 8×(8+x) = 10×(10+6) ⇒ 64 + 8x = 10×16
⇒ 64 + 8x = 160 ⇒ 8x = 96 ⇒ x = 96⁄8 = 12
(b) We have: 10×(10+x) = 12×(12+8) ⇒ 100 + 10x = 12×20
⇒ 100 + 10x = 240 ⇒ 10x = 140 ⇒ x = 140⁄10 = 14
(c) We have: 3×(3+x) = 4×(4+8) ⇒ 9 + 3x = 4×12
⇒ 9 + 3x = 48 ⇒ 3x = 39 ⇒ x = 39⁄3 = 13
Solved example 27.29
Find the value of x in each of the figs given below:
Solution:
(a) We have: 8×(8+x) = 10×(10+6) ⇒ 64 + 8x = 10×16
⇒ 64 + 8x = 160 ⇒ 8x = 96 ⇒ x = 96⁄8 = 12
(b) We have: 10×(10+x) = 12×(12+8) ⇒ 100 + 10x = 12×20
⇒ 100 + 10x = 240 ⇒ 10x = 140 ⇒ x = 140⁄10 = 14
(c) We have: 3×(3+x) = 4×(4+8) ⇒ 9 + 3x = 4×12
⇒ 9 + 3x = 48 ⇒ 3x = 39 ⇒ x = 39⁄3 = 13
In the next section, we will see a discussion on Probability.
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