Saturday, October 29, 2016

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 90o 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 = 90o
9. SP is parallel to RQ. So RSP = 90o
10. PQ is parallel to SR. So SPQ = PQR = 90o
11. Thus we get all angles of the rhombus PQRS in fig.d as 90o. 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
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
In the next section, we will see Similar triangles.


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