Chapter 18.7 - Division of Lines and Triangles - Solved examples

In the previous section we completed the discussion on Parallel lines and Triangle division. We also saw one solved example. In this section, we will see a few more solved examples.

Solved example 18.11

In the parallelogram ABCD in fig.18.36, the line drawn through a point P on AB, parallel to BC, meets AC at Q. The line through Q, parallel to AB meets AD at R. 

Fig.18.36

Prove that:

(i) ^AP⁄_PB = ^AR⁄_RD (ii) ^AP⁄_AB = ^AR⁄_AD

Solution:

Part (i): 

1. Consider the 3 parallel lines AD, PQ and BC. They cut through AC and AB

2. Distances cut on AC and AB are in the same ratio. So we can write: AP/PB = AQ/QC

3. Now consider the other set of 3 parallel lines: DC, RQ and AB. They cut through AC and BC

4. Distances cut on AC and BC are in the same ratio. So we can write: AR/RD = AQ/QC

From (2) and (4), we can write: ^AP⁄_PB = ^AR⁄_RD

Part (ii): 1. Consider the reciprocal of ^AP⁄_AB. It is equal to ^AB⁄_AP . We can write this ^AB⁄_AP in another form:

^AB⁄_AP = ^(AP+BP)⁄_AP = ^AP⁄_AP + ^BP⁄_AP = 1 + ^BP⁄_AP .

2. Consider the reciprocal of ^AR⁄_AD. It is equal to ^AD⁄_AR . We can write this ^AD⁄_AR in another form:

^AD⁄_AR = ^(AR+RD)⁄_AR = ^AR⁄_AR + ^RD⁄_AR = 1 + ^RD⁄_AR .

3. From part (i), we have ^AP⁄_PB = ^AR⁄_RD . Taking reciprocals, we get: ^PB⁄_AP = ^RD⁄_AR .

4. Now consider the results in (1) and (2):

• The second term in the result in (1) is ^BP⁄_AP.

• The second term in the result in (2) is ^RD⁄_AR.

• According to (3) these second terms are equal. It follows that results in (1) and (2) are equal

5. So we get: ^AB⁄_AP = ^AD⁄_AR . Taking reciprocals, we get: ^AP⁄_AB = ^AR⁄_AD 

Solved example 18.12

In the fig.18.37(a)below, the vertex D of the parallelogram ABCD is joined to the midpoint P of side AB. The vertex B is joined to the midpoint Q of side CD. 

Fig.18.37

Prove that these lines PD and QB divide the diagonal AC into three equal parts.

Solution:

1. In fig.a, the lines PD and QB divide the lines into 3 parts: AX, XY and YC. We have to prove that AX = XY = YC

2. ABCD is a parallelogram. So AB = CD

3. P is the midpoint of AB. So AP = BP = ^AB⁄₂ .

4. Q is the midpoint of CD. So DQ = CQ = ^CD⁄₂ .

5. But from (2), AB = CD. So ^AB⁄₂ = ^CD⁄₂.

6. Thus from (3), (4) and (5) we get AP = BP = DQ = CQ = ^AB⁄₂ = ^CD⁄₂ .

7. From (6) we get BP = DQ. Now, BP and DQ are parallel because, they are parts of AB and CD, which are opposite sides of a parallelogram.

8. Thus we can write: BP and DQ are both equal and parallel. So, the quadrilateral PBQD is a parallelogram

9. Since PBQD is a parallelogram, the opposite sides PD and QB will be parallel.

10. Now we give two new companion parallel lines to PD and QB. For that,

• Through C, draw CR parallel to QB

• Through A, draw AS parallel to PD

■ Thus AS, PD, BQ and RC are 4 parallel lines

11. From among the 4, consider the last 3: PD, BQ and RC

12. They cut through CD and AC. The distances cut will be in the same ratio. So we can write: DQ:QC = XY:YC

13. But from (6), DQ = QC. That is., DQ: QC = 1:1

14. So XY:YC will also be equal to 1:1. That means XY = YC

Now, we repeat the above 4 steps from (11), for the left side of the parallelogram ABCD

15. From among the 4 parallel lines, consider the first 3: AS, PD and BQ

16. They cut through AB and AC. The distances cut will be in the same ratio. So we can write: AP:PB = AX: XY

17. But from (6), AP = PB. That is., AP:PB = 1:1

18. So AX:XY will also be equal to 1:1. That means AX=XY
19. From (14) and (18), we get: AX = XY = YC

Solved example 18.13

(i) Prove that, the quadrilateral formed by joining the midpoints of any quadrilateral is a parallelogram. (ii) What if the original quadrilateral is a rectangle? (iii) What if the original quadrilateral is a square?

Solution:

Part (i): In the fig.18.38(a) below, ABCD is any quadrilateral. P, Q, R and S are the midpoints of the sides of the quadrilateral. We have to prove that PQRS is a parallelogram.

Fig.18.38

1. In fig.b, a diagonal BD is drawn. Now consider ΔBCD.

■ Converse of Theorem 18.4 states: In any triangle, the line joining the midpoints of any two sides is parallel to the third side.

2. So RQ is parallel to BD

■ Theorem 18.5 states: The length of a line joining the midpoints of two sides of a triangle is half the length of the third side.

3. So RQ = ^BD⁄₂ .

4. Consider ΔABD

5. Applying theorem 18.4, PS is parallel to BD

6. Applying theorem 18.5, PS = ^BD⁄₂ .

7. From (2) and (5) we get: RQ and PS are parallel

8. From (3) and (6) we get: RQ and PS are equal

9. So, in the inner quadrilateral PQRS, a pair of opposite sides (RQ and PS) are equal and parallel. Then that quadrilateral will be a parallelogram.

[Note that, in the above steps, we used the diagonal BD. We could obtain the same result by using the other diagonal AC also. This is shown in fig.c. In that case, we will get: SR and PQ are equal and parallel]

Part (ii): In fig.18.39(a) below, the original quadrilateral is a rectangle. 

Fig.18.39

In part (i) we have proved that the inner quadrilateral obtained by joining the midpoints of the sides of the outer quadrilateral will be a parallelogram. So, our present inner quadrilateral PQRS is also a parallelogram. But this time, it has some specialities. Let us analyse:

1. Use the diagonal BD as shown in fig.18.39(b). We will get: RQ = SP = ^BD⁄₂ 

2. Use the diagonal AC as shown in fig.18.39(c). We will get PQ = SR = ^AC⁄₂

3. But for a rectangle, the diagonals are equal. So AC = BD ⇒ ^AC⁄₂ = ^BD⁄₂ 

4. So, from (1) and (2), we get: RQ = SP = PQ = SR

5. Thus, the quadrilateral PQRS in fig.18.39, is a 'parallelogram with all the four sides equal'. So it is a Rhombus

Part (iii): In fig.18.40(a) below, the original quadrilateral is a square. 

Fig.18.40

In part (i) we have proved that the inner quadrilateral obtained by joining the midpoints of the sides of the outer quadrilateral will be a parallelogram. So, our present inner quadrilateral PQRS is also a parallelogram. But this time, it has some specialities. Let us analyse:

1. Use the diagonal BD as shown in fig.18.40(b). We will get: RQ = SP = ^BD⁄₂

2. Use the diagonal AC as shown in fig.18.40(c). We will get PQ = SR = ^AC⁄₂

3. But for a square, the diagonals are equal. So AC = BD ⇒ ^AC⁄₂ = ^BD⁄₂

4. So, from (1) and (2), we get: RQ = SP = PQ = SR

5. Thus, the quadrilateral PQRS in fig.18.40, is a 'parallelogram with all the four sides equal'. So it is a Rhombus

• There is more:

6. In a square, the diagonals intersect at right angles. This is indicated by the 90^o angle AOD in fig.d

7. RS is parallel to diagonal AC and RQ is parallel to diagonal BD.

8. But diagonals AC and BD are at right angles to each other. So RS and RQ will also be at right angles to each other. Thus SRQ = 90^o

9. SP is parallel to RQ. So RSP = 90^o

10. PQ is parallel to SR. So SPQ = PQR = 90^o

11. Thus we get all angles of the rhombus PQRS in fig.d as 90^o. So the rhombus PQRS is a square.

Solved example 18.14

In the ΔABC in fig.18.41(a) below, PQ is parallel to AC. QR is parallel to AP. Prove that ^BP⁄_PC = ^BR⁄_RP .

Fig.18.41

Solution:

1. • We need minimum 2 distances to take ratios. • For cutting 2 distances, we need minimum 3 parallel lines. So let us draw a new line BX parallel to PQ through B. This is shown in fig.b.

2. Now we have three parallel lines: BX, PQ, and CA. The distances that they cut on the two lines AB and CB will be in the same ratio. So we can write: ^BP⁄_PC = ^BQ⁄_QA.

3. Draw another new line BY parallel to AP through B. This is shown in fig.c. Now we have another set of 3 parallel lines: BY, RQ and PA

4. The distances that they cut on the two lines AB and BP will be in the same ratio. So we can write:

^BR⁄_RP = ^BQ⁄_QA.

5. The right side of (2) and (4) are the same. So we get: ^BP⁄_PC = ^BR⁄_RP.

Solved example 18.15

In the fig.18.42(a) below, AB and CD are parallel.

Fig.18.42

Prove that AP × PC = BP × PD

Solution:

1. • We need minimum 2 distances to take ratios. • For cutting 2 distances, we need minimum 3 parallel lines. So let us draw a new red line parallel to AB through P. This is shown in fig.b.

2. Now we have three parallel lines: AB, the red line, and CD

3. The distances that they cut on the two lines AD and BC will be in the same ratio. So we can write:

^BP⁄_PC = ^AP⁄_PD. ⇒ AP × PC = BP × PD

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Some more solved examples can be seen in the form of video presentations. The links are given below:
Solved example 18.16     Solved example 18.17

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In the next section, we will see Similar triangles.

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Chapter 18.6 - Triangle division by Medians

In the previous section we saw a new method to find the Circumcentre. In this section, we will see Medians.

We know that, in a triangle, a median is a line drawn from a vertex, to the midpoint of the opposite side. We have seen it before here and here. Let us now draw the three medians of a triangle. 

• Fig.18.32(a) shows a triangle ABC. A median is drawn from vertex C. That is., a line is drawn from the vertex C to the midpoint D of the opposite side BC. 

Fig.18.32

• In fig.b, another median is drawn from the vertex A. 

• And finally, in fig.c, the third median is drawn from vertex B. 

■ Note that, the three medians pass through the same point. We will call this point as ‘G’.

We will now do some calculations with these medians. 

1. Consider any two medians. Let us take those medians from A and B. This is shown in fig.18.33(a) below:

Fig.18.33

2. The ends of the medians are E and F. They are joined by an yellow line.

3. E and F are the midpoints of sides BC and AC. So, by theorem 18.5, EF will be half of AB. That is., EF = ^AB⁄₂

4. Now, there is a smaller triangle ABG inside. Mark the midpoints of it’s sides AG and BG. Let those midpoints be P and Q. This is shown in fig.b

5. Join PQ. Since P and Q are midpoints, by theorem 18.5, PQ will be half of AB. That is., PQ = ^AB⁄₂

6. From (3) and (5), we get PQ = EF

7. We know that PQ is parallel to AB, and EF is also parallel to AB. So PQ and EF are parallel to each other.

8. Thus, in the quadrilateral PQEF shown in fig.18.33(c) below, the opposite sides PQ and EF are equal and parallel. So PQEF is a parallelogram.

Fig.18.33

9. In any parallelogram, the diagonals will bisect each other. So in our parallelogram PQEF, the diagonals PE and FQ will bisect each other. So we can write: • PG = GE  • QG = GF

10. Now let us move back to our original medians AE and BF shown in fig.a. They are now divided at various points P, Q and G. This is shown in fig.d.

11. Consider the median AE. It is divided into three parts AP, PG and GE.

(i) From (4), P is the midpoint of AG. So AP = PG

(ii) From (11.i) and (9), we get: AP = PG = GE. That means, the median AE is divided into three equal parts by points P and G

(iii) On one side of G, there are two equal parts AP and PG, and on the other side, there is one equal part GE.

(iv) So G divides the median AE in the ratio 2:1

The same result of 2:1 ratio can be obtained for the other median BF also. The steps are similar to those from to. But we will write them again:

12. Consider the median BF. It is divided into three parts BQ, QG and GF.

(i) From (4), Q is the midpoint of BG. So BQ = QG

(ii) From (12.i) and (9), we get: BQ = QG = GF. That means, the median BF is divided into three equal parts by points Q and G

(iii) On one side of G, there are two equal parts BQ and QG, and on the other side, there is one equal part GF.

(iv) So G divides the median BF in the ratio 2:1

13. So we find that, G is an important point. It divides the two medians in the ratio 2:1

■ Now, what is this G? Is it very difficult to find the position of G?

Ans: • From fig.18.32(c), G is the point of intersection of the medians of a triangle.

• It is not at all difficult to find the position of G. All we need to do is, to draw any two medians. Their point of intersection is the point ‘G’

14. In fig., we considered a convenient pair of medians. AE and BF. We could take any of the other two possible pairs: • AE and CD  • BF and CD

15. In any case that we take, we will get the same result:

■ The point of intersection of the two medians will divide them both in the ratio 2:1

16. For example, if we take AE and CD, we can write this:

(i) G divides the median AE in the ratio 2:1

(ii) G divides the median CD in the ratio 2:1

From 11, 12  and 16 , we can say that G divides all the three medians in the same ratio 2:1.

We will write it in the form of a theorem:

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Theorem 18.6
The point of intersection G divides all the three medians in the ratio 2:1 measured from the vertex

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Until now, we have been discussing the 'action of parallel lines' in the interior portion of triangles, and also on the sides of the triangles. Now we will have a short discussion on some thing which happens out side a triangle. Fig.18.34(a) shows a triangle ABC. In the fig.b, the sides AC and BC are extended upwards beyond the vertex C.

Fig.18.34

In the fig.18.34(c) below, a line is drawn parallel to the base AB, cutting through the extended portions. Let it intersect the extensions at P and Q.

Fig.18.34

A red line parallel to AB is drawn through C. This is shown in fig.d. We do not need the portions beyond P and Q. So they are trimmed.

1. Thus, in fig.d, we have three parallel lines: • AB, • 'the red line' and • PQ, 

2. Those parallel lines cut through the lines AQ and BP. The distances cut are in the same ratio (Theorem 18.1). So we have: AC:QC = BC:PC ⇒ ^AC⁄_QC  = ^BC⁄_PC .

3. Consider a different ratio: ^PB⁄_PC .

^PB⁄_PC = ^{(PC + BC)}⁄_PC (∵ PB = PC + BC)

⇒ ^PB⁄_PC = ^PC⁄_PC + ^BC⁄_PC ⇒ ^PB⁄_PC = 1 + ^BC⁄_PC . 

4. Similarly, consider:  ^AQ⁄_CQ .

^AQ⁄_CQ = ^{(CQ + AC)}⁄_CQ (∵ AQ = CQ + AC)

⇒ ^AQ⁄_CQ = ^CQ⁄_CQ + ^AC⁄_CQ ⇒ ^AQ⁄_CQ = 1 + ^AC⁄_CQ .

5. Let us rewrite the results in (3) and (4):

From (3) we have: ^PB⁄_PC  = 1 + ^BC⁄_PC .

From (4) we have: ^AQ⁄_CQ = 1 + ^AC⁄_CQ . 

6. The last terms in the above two equations are the same. This we know from (2)

7. So (3) and (4) are equal. That is., ^PB⁄_PC  =  ^AQ⁄_CQ . 

■ We can write a summary of the above discussion as follows:

• Extend the sides of a triangle

• Consider one side. Take the ratio: ^(Extension)⁄_{(Total length)}

• Consider the other side. Take the ratio: ^(Extension)⁄_{(Total length)}

• Both the ratios will be equal

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Now we will see a solved example
Solved example 18.10

In the fig.13.35 below, ABC is a right angled triangle. D is the midpoint of the hypotenuse AC. DE is drawn perpendicular to AB.

Fig.18.35

(i) Prove that DE is half of BC (ii) Prove that, in the larger ⊿ABC, the distances from D to all the vertices are equal (iii) Prove that D is the circumcentre of ⊿ABC

Solution:

Part (i): 1. DE is parallel to BC. Because, both DE and BC are perpendicular to AB. Also, D is the midpoint of AC. 
2. Theorem 18.4 states: In any triangle, the line drawn parallel to one side, passing through the midpoint of another side, meets the third side also at it's midpoint.
■ So E is the midpoint of AB
3. Now we apply theorem 18.5: The length of a line joining the midpoints of two sides of a triangle is half the length of the third side.
■ So DE is half of BC

Part (ii): The distances to the three vertices are DA, DB and DC. We have to prove that, these three distances are equal.

1. We already know that DA = DC. Because D is the midpoint of AC

Now we will prove that DA = DB. The proof is as follows:

2. In fig.18.35(b) above, consider the triangles: ⊿AED and ⊿BED

• ED = ED (The common side)

• AE = BE (since E is the midpoint of AB)

• ∠AED and ∠BED are colinear angles, and one of them (∠AED) ie 90^o. So ∠BED = 180 – 90 = 90^o

• Thus, ∠AED = ∠BED = 90^o

3. Thus we have two sides and included angle same in both the triangles. It is a case of SAS congruence. The two triangles ⊿AED and ⊿BED are equal.

• Let us write the correspondence:

• ED is the common side. The 90^o is at E. So E↔E and D↔D

• The remaining vertices are A and B. So A↔B

• We can write: The correspondence of vertices is: E↔E, D↔D and A↔B

• From the above, we can pick each corner and write the correspondence of sides: ED↔ED, EA↔EB and DA↔DB

• From the last one DA↔DB, we can write DA= DB

4. From [Part (ii) 1], we have: DA = DC

• From [Part (ii) 3], we have DA = DB

• So we can write DA = DB = DC

5. Thus, the distances from D to all the vertices of ⊿ABC are equal

Part (iii): 
• If D is the circumcentre, there exists a circle, with D as it's centre, and with the vertices A, B and C lying on it.

• Indeed there exists such a circle because in part (ii), we proved that A, B and C are at equidistance from D. SO D is the circumcentre of ABC

We can prove this in another way also:

1. Draw the perpendicular bisector of BC as shown in fig.c below

Fig.18.35

2. Let it bisect BC at F. Since, it is a bisector, BF is half of BC

3. Now, using theorem 18.5, DE is also half of BC

4. That means., D and F, are at the same perpendicular distance from the side AB. Also, the perpendicular bisector through F will be parallel to AB

5. So, if we extend the perpendicular bisector, it will pass through D. See fig.d

6. We already know that DE is the perpendicular bisector of AB (since it is perpendicular to AB, and E is the midpoint of AB)

7. So we have the perpendicular bisectors of two sides intersecting at D. Thus D is the circumcentre

An even simpler method:

1. The perpendicular bisector of AB, already passes through D

2. We need one more perpendicular bisector.

3. Why not take the one of the hypotenuse AC. It will pass through D itself.

4. So we get two perpendicular bisectors, and both of them intersect at D. Thus D is the circumcentre

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

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Chapter 18.4 - Division of Triangles using Parallel lines

In the previous section we completed the discussion on Division of lines. In this section, we will see Triangle division.

Consider the triangle ABC in fig.18.25(a) below. A red line is drawn parallel to the base AB. This red line intersect the sides AC and BC at P and Q. 

Fig.18.25

At a first glance there is nothing special about this fig(a). But some important details are hidden in it. Let us bring them out:

Draw two more parallel red lines. One through the apex C, and another through the base AB. This is shown in the fig(b). Now we have three parallel red lines. They cut through two lines AC and BC. So the distances cut will be in the same ratio. So we can write: AP:CP = BQ:CQ

This gives us a method to draw a line parallel to the base of any triangle. Let us see an example:

Solved example 18.9

In the triangle ABC in the fig.18.26(a) below, AC = 10 cm, BC = 15 cm and AP = 4 cm. Draw a line parallel to AB through P

Fig.18.26

Solution:

• To draw a parallel to AB through P, we need the perpendicular distance from P to AB. But it is not given.

• What we have, is the distance AP. Let us check whether there is any relation between this distance and the total distance AC:

• We have: ^AP⁄_AC = ⁴⁄₁₀ = ²⁄₅ ⇒ AP = ²⁄₅ × AC.

• That means., AP is two fifths of AC. In other words, if we divide AC into 5 equal parts, AP will take up two such parts. It follows that, CP will take up the remaining three parts.

• Thus we can say: P divides AC in such a way that AP: CP = 2:3

• Now, we need a point Q on BC, which will divide BC in such a way that BQ:CQ = 2:3. Let us find this 'Q':

• We have: BQ:CQ = 2:3. So if BC is divided into 5 equal parts, BQ should take up 2 such parts. In other words, BQ must be two fifths of BC

• Thus we get: BQ = ²⁄₅ × BC = ²⁄₅ × 15 = 6 cm

• With B as centre, and 6 cm as radius, draw an arc. This is shown in fig(b). This arc will cut BC at the required point Q

• Join P and Q. The Line PQ will be parallel to AB

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In the above problem, we effectively used theorem 18.1 to draw the required parallel line. We found that P divides AC in the ratio 2:3. 

■ In fact we can use the method for any ratio. Because, we have learned how to divide any line in any given ratio.

■ So what if P is the exact midpoint of AB?

• Then the calculations steps are even more easier. The ratio is a simple 1:1. All we have to do is to draw a perpendicular bisector of the other side BC. This bisector will give us the point Q, and we can draw PQ

• From this we get some new information. We can write it in the form of a theorem. We will write it in steps.

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Theorem 18.4

1. We have a triangle ABC with base AB, and two sides AC and BC

2. We mark the midpoint of AC as P, and the midpoint of BC as Q

3. Then we join PQ. This PQ will be parallel to the base AB

We can write the converse also:

1. We have a triangle ABC with base AB, and two sides AC and BC

2. We mark the midpoint of AC as P

3. Then we draw a line through P, parallel to the base AB. This parallel line meets the other side BC at Q

4. Q will be the midpoint of BC
• Note that, in the triangle, any side can be taken as the 'base'. The 'two sides' will change accordingly.
■ The above steps can be written in just one line as:
In any triangle, the line drawn parallel to one side, passing through the midpoint of another side, meets the third side also at it's midpoint

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Let us join the midpoints of all the 3 sides of a triangle. This is shown in the fig.18.27(a) below:

Fig.18.27

1. P is the midpoint of AC; Q is the midpoint of BC; R is the midpoint of AB

2. Joining P, Q and R, we get an inner triangle PQR

3. Also we have: PQ parallel to AB (∵ P is midpoint of AC & Q is midpoint of BC, using theorem 18.4)

4.QR parallel to AC (∵ Q is midpoint of BC & R is midpoint of AB)

5. PR parallel to BC (∵P is midpoint of AC & R is midpoint of AB)

• Based on the above 5 points on fig.18.27(a), we can derive some very interesting results:

6. We have AC parallel to QR ⇒ AP parallel to QR

7. Also we have AB parallel to PQ ⇒ AR parallel to PQ

• Let us take out this portion ARQP for analysis. It is shown separately in fig.18.27(b)

8. We have two parallel lines PQ and AR cut by a transversal PR. So ∠ARP = ∠RPQ 
• They are alternate interior angles as shown in fig.18.27(c) below:

Fig.18.27

• The equal angles ∠ARP and ∠RPQ are shown in green colour

9. We have two parallel lines AP and QR cut by a transversal PR. So ∠APR = ∠PRQ
• They are alternate interior angles as shown in fig.18.27(c) above.
• The equal angles ∠APR and ∠PRQ are shown in white colour

10. Now consider the line PR. What are the angles at it's end, from a point of view of ΔPQR?

Ans: The angles are ∠RPQ (green) and ∠PRQ (white)

11. What are the angles at it's end, from a point of view of ΔAPR?

Ans: The angles are ∠APR (white) and ∠ARP (green)

12. But from (9), ∠APR (white) = ∠PRQ (white), and from (8), ∠ARP (green) = ∠RPQ (green)

• Now ask the question (11) again:

13. What are the angles at the ends of line PR, from a point of view of ΔAPR?

Ans: The angles are ∠PRQ (white) and ∠RPQ (green)

14. The answers in (10) and (13) are the same. Line PR has green angle and white angle at it's ends in both the triangles. Thus. we have a line PR, and the angles at it's ends, present in both ΔPQR and ΔAPR

15. It is a case of ASA congruence (Angle, Side, Angle). The two triangles are equal. That is: ΔPQR = ΔAPR

16. Also, both the pairs of opposite sides are parallel. So ARQP is a parallelogram

17. We proved (15) and (16), by taking out the portion ARQP 

18. In the same way, by taking out the portion PRBQ, we can prove:

(i)ΔPQR = ΔQRB and (ii) PRBQ is a parallelogram

19. In the same way, by taking out the portion PRQC, we can prove:

(i) ΔPQR = ΔPQC and (ii) PRQC is a parallelogram

20. From (15), (18.i) and (19.i), we can write:

ΔPQR = ΔAPR = ΔQRB = ΔPQC

■ That is., all the four inner triangles are equal.

Another interesting result:

21. Consider the non-isosceles trapezium ABQP

(i) AR = PQ. [since from (16), ARQP is a parallelogram, and AR and PQ are it's opposite sides]
(ii) Also, from (1), R is the midpoint of AB

• It follows that, PQ = AR = RB. That means, PQ is half of AB

22. In a similar way, by considering the non-isosceles trapezium BCPR, we get: PR is half of BC

23. In a similar way, by considering the non-isosceles trapezium ACQR, we get: QR is half of AC

• We can write the above findings (21), (22) and (23) in the form of a theorem. In fact it is an extension of theorem 18.4

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Theorem 18.5

1. We have a triangle ABC with base AB, and two sides AC and BC

2. We mark the midpoint of AC as P, and the midpoint of BC as Q

3. Then we join PQ. This PQ will be parallel to the base AB

4. Also PQ will be half of AB

We can write the converse also:

1. We have a triangle ABC with base AB, and two sides AC and BC

2. We mark the midpoint of AC as P

3. Then we draw a line through P, parallel to the base AB. This parallel line meets the other side BC at Q

4. Q will be the midpoint of BC

5. Also PQ will be half of AB
• Note that, in the triangle, any side can be taken as the 'base'. The 'two sides' will change accordingly.
■ The above steps can be written in just one line as:
The length of the line joining the midpoints of two sides of a triangle is half the length of the third side

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We will now see a solved example based on the above discussion
Solved example 18.9
Fig.18.28 below shows a right angled triangle ABC. The midpoint of AC is marked as D. A perpendicular DE is dropped from D to the base AB. (i) Calculate the side CB of the larger triangle ABC. (ii) Calculate all the sides AE, AD and ED of the smaller triangle AED

Fig.18.28

Solution: 
Part (i): 1. ABC is a right angled triangle, as indicated by the 90^o at the corner B. We can simply apply Pythagoras theorem to ⊿ABC:
CB² = 10² - 8² ⇒ CB² = 100 - 64 ⇒ CB² = 36 ⇒ CB = √36 = 6 cm
Part(ii): 1. DE is parallel to BC. Because, both DE and BC are perpendicular to AB. Also, given that D is the midpoint of AC. 

2. Theorem 18.4 states: In any triangle, the line drawn parallel to ones side, passing through the midpoint of another side, meets the third side also at it's midpoint.

3. So E is the midpoint of AB. Thus we get AE = ⁸⁄₂ = 4 cm

4. Now we apply theorem 18.5: The length of a line joining the midpoints of two sides of a triangle is half the length of the third side.

5. So DE is half of BC. Thus we get DE =  ⁶⁄₂ = 3 cm

6. Given that D is the midpoint of AC. So DE =  ¹⁰⁄₂ = 5 cm

7. (3), (5) and (6) gives the answer to part (ii)

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In the next section, we will see more details about Triangle Division.

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Chapter 14.7 - Division of Triangles - Solved examples

In the previous section we completed the discussion on the division of triangles. We also saw some solved examples. In this section we will see some more solved examples that demonstrate the topics that we have discussed in this chapter as a whole.

Solved example 14.20
ABCD is a trapezium. It’s diagonals AC and BD meet at O. Prove that the magenta coloured ΔAOD and the red coloured ΔBOC have the same area.

Fig.14.33

Solution:
1. In any trapezium, two opposite sides are parallel. In the fig., the parallel sides are AB and CD
2. In the fig., ΔABD and ΔABC have the same area. [since they are triangles between the same parallels, and they have the same base AB]
3. So we can write: ar (ABD) = ar (ABC)
4. But [ar (ABD) = ar (ABO) + ar (AOD)] and [ar (ABC) = ar (ABO) + ar (BOC)]
5. Substituting these values in 3, we get:
[ar (ABO) + ar (AOD)] = [ar (ABO) + ar (BOC)] That is., the yellow ΔABO is common to both
6. ar (ABO) present on both sides will cancel out. So we get:
ar (AOD) = ar (BOC)

Solved example 14.21

In the previous example, what is the total area of the trapezium ABCD, if the area of the blue triangle is 4 cm² and yellow triangle is 9 cm²
Solution:
1. In the previous example, we have already proved that ar (AOD) = ar (BOC). Let each be equal to x. That is.,
2. Let ar (AOD) = ar (BOC) = x cm²
3. From fig.14.33(b) we get • ar (AOD) =  ¹⁄₂ × AO × h₁ and   • ar (COD) = ¹⁄₂ × CO × h₁

4. Also we get • ar (AOB) =  ¹⁄₂ × AO × h₂ and   • ar (COB) = ¹⁄₂ × CO × h₂ 

5. From (3) and (4) we get

6. So we get, total area of the trapezium = 9 + 4 + x + x = 9 + 4 + 6 + 6 = 25 cm² 

Solved example 14.22
In fig.14.34(a), CD is a median of the ABC. This median CD is divided at E in such a way that CE : DE = 2:1. Prove that area of each triangle in fig.13.34(b) is one third of the whole area of ΔABC

Fig.14.34

Solution:
1. Consider ΔADC. It is split into two triangles: ΔADE and ΔACE.
2. Given that CE : DE = 2 :1. So ar (ACE) : ar (ADE) = 2 :1
3. That means ar (ACE) = 2 × ar (ADE)
4. In a similar way, ar (BCE) = 2 × ar (BDE)

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5. Consider ΔABE. It is split into two triangles: ΔADE and ΔBDE
6. CD is a median. So AD : BD = 1 : 1. So ar (ADE) = ar (BDE)
7. So we can put ar (ADE) in the place of ar (BDE) in (4)
8. We get ar (BCE) = 2 × ar (ADE)
9. Compare (3) and (8). The right sides are the same. So left sides also must be equal
10. We get ar (ACE) = ar (BCE)

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11. Take the sum of two triangles: ΔADE and ΔBDE:
12. We get ar (ADE) + ar (BDE) = ar (ABE). But from (6), ar (BDE) = ar (ADE)
13. ∴ 2 × ar (ADE) = ar (ABE) - - -(6)
14. Comparing the above with (3) we get ar (ACE) = ar (ABE)
15. Comparing (14) and (10) we get  ar (ACE) = ar (ABE) = ar (BCE)
16. The sum of the 3 triangles in (15) is the total area ar (ABC). Each of the three are equal. That means each triangle is equal to one third of the total area

Solved example 14.23
In fig.14.35(a), ABCD is a parallelogram. AB is extended to any point P. Line AQ is drawn through A, parallel to PC. AQ meets CB produced at Q. Parallelogram BQRP is completed by drawing QR and RP. Prove that ar (ABCD) = ar (BQRP). [Hint: Draw the diagonals of the parallelograms] 

Fig.14.35

Solution:
1. The diagonals AC and PQ are added to the fig. in 14.25(a). The modified fig. is shown in (b)
2. Given that PC is parallel to AQ
3. So AQC and AQP are two triangles with the same base, and between same parallels. 
4. Thus, they have the same area. That is., ar (AQC) = ar (AQP)
5. Let us split the above two areas:
• ar (AQC) = ar (ABC) + ar (AQB)
• ar (AQP) = ar (BQP) + ar (AQB)
6. Let us equate as in (4): [ar (ABC) + ar (AQB)] = [ar (BQP) + ar (AQB)]
7. ar (AQB) is common. It will cancel out. 
8. So we get ar (ABC) = ar (BQP)
9. In (8) above, each is half of the corresponding parallelogram. (∵ AC and PQ are diagonals)
10. Doubling each will give the corresponding parallelogram. So we get ar (ABCD) = ar (BQRP)

Solved example 14.24
Prove that the perpendiculars drawn from any point on the angle bisector to the sides are equal
Solution:
1. The diagram for this example is given in fig.14.36(a) below:

Fig.14.36

We have:
• an ∠BCA, • it's bisector BG, • 'any point' D on the bisector, • perpendiculars DE and DF from D, to the sides
2. We have to prove that DF and DE are equal
3. We have proved the above equality when we discussed Theorem 14.7
4. In fact, the fig. 14.36(a) given above is the same fig.14.28(b) that we saw when we discussed the theorem
5. There we proved that DF and DE are equal, and so, indicated each of them as 'h'. The same steps can be followed in this example also

Solved example 14.25
In the fig.14.36(b), ABCD is a rectangle of length 14 cm, and width 6 cm. E is the midpoint of BC. F is the mid point of AE. Find the areas of ΔABF and ΔEBF
Solution:
1. Consider ΔABE. Base AB = 14 cm. Height BE = 6/2 = 3 cm (∵ E is the midpoint of BC)
2. So ar (ABE) = ¹⁄₂ × 14 × 3 = 21 cm²
3. F is the midpoint of AE. So BF is a median of ΔABE
4. So ar (ABF) = ar (EBF) [By Theorem 14.5]
5. If the two areas are equal, each must be exact half of the total ar (ABE)
6. But ar (ABE) = 21 cm²
7. So we get ar (ABF) = ar (EBF) = 21/2 = 10.5 cm²

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This completes the discussion on Triangles. In the next section we will see 'Pairs of Equations'.

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