Chapter 18 - Three parallel lines cut by Transversal

In the previous section we completed the discussion on Circles and chords. In this section, we will discuss about Parallel lines.

In fig.18.1 below, 3 parallel lines Line A, Line B and Line C are drawn in red colour. Across these parallel lines, Line 0 is drawn in a perpendicular direction. Lines (1), (2) and (3) are drawn in random directions. These transverse lines are drawn in cyan colour.

Fig.18.1

The red parallel lines cut the cyan lines at A₀ , B₀ , A₁ , A₂ etc., For our present discussion, we want the distances between these points of intersection. Those distances are measured and marked in fig.18.2 shown below:

Fig.18.2

From the fig, we can see that the distance A₀B₀= 2.00 cm, B₀C₀ = 4.00 cm, A₂B₂ = 2.38 cm etc.,

---

Note that, these distances given in fig.18.2 are the distances between the red parallel lines, measured along the cyan transverse lines. The shortest distances will of course be when measured along the perpendicular line, which is Line 0.

---

Once we mark all the distances, we can begin the calculations.

1. Take the ratio A₀B₀ : B₀C₀ .

• It will be equal to 2:4.

• Reducing to the lowest term, 2:4 is equal to 1:2 [∵ 4.00 ÷ 2.00 = 2]

• That is., A₀B₀ : B₀C₀ = 1:2

2. Now take the corresponding ratio along Line 1.

• We get A₁B₁ : B₁C₁ = 2.03 : 4.06

• Reducing to the lowest term, 2.03 : 4.06 is equal to 1:2 [∵ 4.06 ÷ 2.03 = 2]

• That is., A₁B₁ : B₁C₁ = 1:2

3. Now take the corresponding ratio along Line 2.

• We get A₂B₂ : B₂C₂ = 2.38 : 4.76

• Reducing to the lowest term, 2.38 : 4.76 is equal to 1:2 [∵ 4.76 ÷ 2.38 = 2]

• That is., A₂B₂ : B₂C₂ = 1:2

4. Finally, take the corresponding ratio along the last Line 3.

• We get A₃B₃ : B₃C₃ = 2.10 : 4.20

• Reducing to the lowest term, 2.10 : 4.20 is equal to 1:2 [∵ 4.20 ÷ 2.10 = 2]

• That is., A₃B₃ : B₃C₃ = 1:2

■ So we get the same ratio 1:2 in all the four cases. In fact, we will get the same ratio for any number of transverse cyan lines that we draw, in any direction that we like.

■ That means, in the fig.18.2, the 'distance between Line B and Line C' will always be twice the 'distance between Line A and Line B', in what ever direction we take the measurements. The ratio is a constant.

■ For the three red parallel lines in fig.18.1, the constant ratio is 1:2. 

■ We will write a general form applicable to any ratio:

• Let the constant ratio be m:n

• m:n can be written as 1:(ⁿ⁄_m)

• So the 'distance between Line B and Line C' will always be (ⁿ⁄_m) times the 'distance between Line A and Line B', in what ever direction we take the measurements.

---

This property can be used to determine unknown quantities. Let us see a quick example:

In fig.18.3(a) below, AB, EF and CD are parallel. Determine the length BF

Fig.18.3

Solution:

• Given that AB, EF and CD are parallel. So they can be considered as the red parallel lines that we saw in fig.18.1. This is shown in fig.18.3(b). 

• AD and BC are transverse lines that cross the parallel lines. What ever be the direction of the transverse lines, the ratio of the distances will be a constant. 

• So we can write: AE:ED = BF:FC = 2:5

• 2:5 can be written as 1: ⁵⁄₂ .
• Thus we get: AE:ED = BF:FC = 2:5 = 1: ⁵⁄₂
• That means: ED is ⁵⁄₂ times AE.

• FC will also be ⁵⁄₂ times BF.

• So 4.45 = ⁵⁄₂ × BF

• Thus BF = 4.45 × ²⁄₅ = 1.78
Another method:
■ We have AE:ED = 2:5. This gives us the following information:
• If the total length from A to D is divided into 7 equal parts, AE will take up 2 parts, and DE will take up 5 parts.
• BF:FC is also the same 2:5
• So, if the total length from B to C is divided into 7 equal parts, BF will take up 2 parts, and FC will take up 5 parts.
• Thus we get: 5 parts out of '7 equal parts of BC' = 4.45
• That is: BC × ⁵⁄₇ = 4.45 ⇒ BC = 4.45 × ⁷⁄₅ = 6.23
• Now, 2 parts out of 7 equal parts of BC is BF
• That is: BC × ²⁄₇ = BF ⇒ BF = 6.23 × ²⁄₇ = 1.78

---

So we find that the ratio is a constant. We will now see the official proof:

1. Consider fig.18.4(a) below. We have three parallel red lines and two transverse cyan lines. They intersect at A, B, C etc.,

Fig.18.4

2. Consider fig.18.4(b). We have a triangle in yellow colour: ΔADF.

• Consider this triangle to have AD as it's base, and F as the apex. So FE is a line from the apex to the base

• FE splits ΔADF into two: ΔAEF and ΔDEF

[For convenience, let us denote the areas by 'ar'. So:

• ar(AEF) will denote area of ΔAEF

• ar(DEF) will denote area of ΔDEF and so on...]

3. We have: ar(AEF) : ar(DEF) = AE : DE [This we know from Theorem 14.6 which we discussed based on fig.14.26]

■ The same set of calculations can be applied to the green triangle in fig.(c). Let us write them:

4. Consider fig.18.4(c). We have a triangle in green colour: ΔBCE.

• Consider this triangle to have BC as it's base, and E as the apex. So EF is a line from the apex to the base.

• EF splits ΔBCE into two: ΔBEF and ΔCEF

• Using theorem 14.6, we have: ar(BEF) : ar(CEF) = BF : CF

■ Now we look at the two triangles from another view point:

First we put them together as in fig.(d)

5.Consider the portion below line EF. There, we have two triangles: ΔEFA and ΔEFB

• Both the triangles have the same base EF

• Apex of both the triangles lie on the same line AB

• This AB is parallel to the base EF

From the above three points, we have: ar(AEF) = ar(BEF) [This we know from Theorem 14.1]

■ The same set of calculations can be applied to the portion above line EF. Let us write them:

6. There, we have two triangles: ΔEFC and ΔEFD

• Both the triangles have the same base EF

• Apex of both the triangles lie on the same line CD

• This CD is parallel to the base EF

From the above three points, using theorem 14.1, we have: ar(CEF) = ar(DEF)

■ The main set of calculations are over. We will take out the results that we require:

7. From (3) we have: ar(AEF) : ar(DEF) = AE : DE

8. From (4) we have: ar(BEF) : ar(CEF) = BF : CF

9. From (5) we have: ar(AEF) = ar(BEF)

10. From (6) we have: ar(CEF) = ar(DEF)

11. • In (9), we have another value for ar(AEF), which is ar(BEF)

• In (10), we have another value for ar(DEF), which is ar(CEF)     

12. We will substitute these new values in (7). We get: ar(BEF) : ar(CEF) = AE : DE

13. But in (13), we have:  ar(BEF) : ar(CEF) = BF : CF

14. Comparing (7), (12) and (13), we get: ar(AEF) : ar(DEF) = AE : DE = ar(BEF) : ar(CEF) = BF : CF

15. From (14) we get: AE : DE = BF : CF

That means, the ratios are same.

---

So we proved that the ratio is a constant for any 3 red parallel lines. In the next section, we will see 4 red Parallel lines.

PREVIOUS      CONTENTS       NEXT

                        Copyright©2016 High school Maths lessons. blogspot.in - All Rights Reserved