GSEB Class 10 Maths Solutions Chapter 6 Triangle Exercise 6.2

Question 1. In figure (i) and (ii). DE || BC. Find EC in (i) and AD in (ii).

Answer:
(i) In \( \triangle ABC \), since \( DE \parallel BC \), by the Basic Proportionality Theorem (BPT), we can write:
\( \frac{AD}{DB} = \frac{AE}{EC} \)
Substituting the given values from the figure:
\( \frac{1.5}{3} = \frac{1}{EC} \)
To find \( EC \), we can cross-multiply:
\( 1.5 \times EC = 3 \times 1 \)
\( 1.5 \times EC = 3 \)
\( EC = \frac{3}{1.5} \)
\( EC = 2 \) cm
(ii) In \( \triangle ABC \), since \( DE \parallel BC \), by the Basic Proportionality Theorem (BPT), we can write:
\( \frac{AD}{DB} = \frac{AE}{EC} \)
Substituting the given values from the figure:
\( \frac{AD}{1.8} = \frac{7.2}{5.4} \)
To find \( AD \), we can cross-multiply:
\( AD = \frac{7.2 \times 1.8}{5.4} \)
\( AD = \frac{12.96}{5.4} \)
\( AD = 2.4 \) cm
In simple words: We used a rule called BPT because the line DE is parallel to BC. This rule lets us set up ratios of the sides. Then, we just put in the numbers we know and solve to find the missing length.

Exam Tip: Remember to clearly state "By BPT" when applying the Basic Proportionality Theorem. Make sure to use the correct corresponding sides in the ratio.

Question 2. E and F are points on the sides PQ and PR respectively of a \( \triangle PQR \). For each of the following cases, state whether EF || QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, and FR = 2.4 cm.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, and RF = 9 cm.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, and PF = 0.36 cm.
Answer:
(i) We are given:
PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
First, calculate the ratio \( \frac{PE}{EQ} \):
\( \frac{PE}{EQ} = \frac{3.9}{3} = \frac{39}{30} = \frac{13}{10} = 1.3 \) .........(1)
Next, calculate the ratio \( \frac{PF}{FR} \):
\( \frac{PF}{FR} = \frac{3.6}{2.4} = \frac{36}{24} = \frac{3}{2} = 1.5 \) .........(2)
From equations (1) and (2), we see that:
\( \frac{PE}{EQ} \neq \frac{PF}{FR} \)
Therefore, by the converse of BPT, EF is not parallel to QR.

(ii) We are given:
PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
First, calculate the ratio \( \frac{PE}{EQ} \):
\( \frac{PE}{EQ} = \frac{4}{4.5} = \frac{40}{45} = \frac{8}{9} \) .........(1)
Next, calculate the ratio \( \frac{PF}{FR} \):
\( \frac{PF}{FR} = \frac{8}{9} \) .........(2)
From equations (1) and (2), we see that:
\( \frac{PE}{EQ} = \frac{PF}{FR} \)
Therefore, by the converse of BPT, EF || QR.

(iii) We are given:
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
First, find EQ and FR:
\( EQ = PQ - PE = 1.28 - 0.18 = 1.10 \) cm
\( FR = PR - PF = 2.56 - 0.36 = 2.20 \) cm
Next, calculate the ratio \( \frac{PE}{EQ} \):
\( \frac{PE}{EQ} = \frac{0.18}{1.10} = \frac{18}{110} = \frac{9}{55} \) .........(1)
Then, calculate the ratio \( \frac{PF}{FR} \):
\( \frac{PF}{FR} = \frac{0.36}{2.20} = \frac{36}{220} = \frac{9}{55} \) .........(2)
From equations (1) and (2), we see that:
\( \frac{PE}{EQ} = \frac{PF}{FR} \)
Therefore, by the converse of BPT, EF || QR.
In simple words: To check if a line inside a triangle is parallel to one of its sides, we use the converse of BPT. This means we compare the ratios of the segments on the other two sides. If the ratios are equal, the lines are parallel; if not, they are not parallel.

Exam Tip: Always remember to subtract to find the segment lengths (like EQ = PQ - PE) when the total side length is given. Clearly state your conclusion based on the converse of BPT.

Question 3. In figure, if LM || CB and LN || CD, prove that \( \frac{AM}{AB} = \frac{AN}{AD} \).

Answer:
Given:
In \( \triangle ABC \), \( LM \parallel CB \)
In \( \triangle ACD \), \( LN \parallel CD \)
To prove:
\( \frac{AM}{AB} = \frac{AN}{AD} \)
Proof:
1. In \( \triangle ABC \), since \( LM \parallel CB \), by the Basic Proportionality Theorem (BPT), we have:
\( \frac{AM}{MB} = \frac{AL}{LC} \) .........(1)
2. In \( \triangle ACD \), since \( LN \parallel CD \), by the Basic Proportionality Theorem (BPT), we have:
\( \frac{AN}{ND} = \frac{AL}{LC} \) .........(2)
3. From equation (1) and (2), we can compare the ratios, since both are equal to \( \frac{AL}{LC} \):
\( \frac{AM}{MB} = \frac{AN}{ND} \)
4. Now, we use the property of Invertendo (taking reciprocals):
\( \frac{MB}{AM} = \frac{ND}{AN} \)
5. Add 1 to both sides of the equation:
\( \frac{MB}{AM} + 1 = \frac{ND}{AN} + 1 \)
\( \frac{MB + AM}{AM} = \frac{ND + AN}{AN} \)
6. Since \( MB + AM = AB \) and \( ND + AN = AD \), substitute these into the equation:
\( \frac{AB}{AM} = \frac{AD}{AN} \)
7. Finally, apply Invertendo again to get the desired result:
\( \frac{AM}{AB} = \frac{AN}{AD} \)
Hence, proved.
In simple words: We used BPT in two different triangles because of the parallel lines. This gave us two ratio equations. By comparing these equations, we found a common relationship. Then, we did a few algebraic steps like taking reciprocals and adding 1 to both sides to change the form of the ratio until we got what we needed to prove.

Exam Tip: When using BPT to prove ratios involving the full side (like AB or AD), remember to use the Invertendo property and add 1 to both sides to convert partial ratios (like AM/MB) into full ratios (like AM/AB).

Question 4. In figure, DE || AC, and DF || AE prove that \( \frac{BF}{FE} = \frac{BE}{EC} \).

Answer:
Given:
In the figure, \( DE \parallel AC \) and \( DF \parallel AE \).
To prove:
\( \frac{BF}{FE} = \frac{BE}{EC} \)
Proof:
1. Consider \( \triangle BAE \). Since \( DF \parallel AE \), by the Basic Proportionality Theorem (BPT), we have:
\( \frac{BD}{DA} = \frac{BF}{FE} \) .........(1)
2. Consider \( \triangle BAC \). Since \( DE \parallel AC \), by the Basic Proportionality Theorem (BPT), we have:
\( \frac{BD}{DA} = \frac{BE}{EC} \) .........(2)
3. From equation (1) and equation (2), since both ratios are equal to \( \frac{BD}{DA} \), we can equate them:
\( \frac{BF}{FE} = \frac{BE}{EC} \)
Hence, proved.
In simple words: We used BPT two times, once in a smaller triangle (\( \triangle BAE \)) and once in the larger triangle (\( \triangle BAC \)), because we were given two sets of parallel lines. Both applications gave us a ratio involving \( \frac{BD}{DA} \). By setting these two ratios equal to each other, we directly proved the required relationship between the sides.

Exam Tip: When dealing with nested triangles and parallel lines, break down the problem. Apply BPT to the relevant smaller triangle first, then to the larger triangle that contains it, and look for a common ratio to connect your findings.

Question 5. In figure DE || OQ and DF || OR. Show that EF || QR. (Foreign 2008)

Answer:
Given:
In \( \triangle POQ \), \( DE \parallel OQ \)
In \( \triangle POR \), \( DF \parallel OR \)
To show:
\( EF \parallel QR \)
Solution:
1. Consider \( \triangle POQ \). Since \( DE \parallel OQ \), by the Basic Proportionality Theorem (BPT), we have:
\( \frac{PE}{EQ} = \frac{PD}{DO} \) .........(1)
2. Consider \( \triangle POR \). Since \( DF \parallel OR \), by the Basic Proportionality Theorem (BPT), we have:
\( \frac{PF}{FR} = \frac{PD}{DO} \) .........(2)
3. From equation (1) and equation (2), since both ratios are equal to \( \frac{PD}{DO} \), we can equate them:
\( \frac{PE}{EQ} = \frac{PF}{FR} \)
4. Now, consider \( \triangle PQR \). We have shown that the line EF divides the sides PQ and PR in the same ratio. Therefore, by the converse of the Basic Proportionality Theorem (BPT), EF || QR.
Hence, proved.
In simple words: We used the BPT twice, once in the triangle POQ and again in triangle POR, because DE is parallel to OQ and DF is parallel to OR. This gave us two sets of equal ratios. Since both ratios shared a common part, we could say that \( \frac{PE}{EQ} \) equals \( \frac{PF}{FR} \). Then, using the converse of BPT in the big triangle PQR, we concluded that EF must be parallel to QR.

Exam Tip: This question is a classic application of BPT and its converse. Remember to apply BPT to the two smaller triangles first to establish a common ratio, and then use the converse of BPT in the larger triangle to prove parallelism.

Question 6. In figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. (CBSE 2007)

Answer:
Given:
Points A, B, and C are located on sides OP, OQ, and OR respectively, of \( \triangle PQR \).
\( AB \parallel PQ \)
\( AC \parallel PR \)
To show:
\( BC \parallel QR \)
Solution:
1. Consider \( \triangle POQ \). Since \( AB \parallel PQ \) (given), by the Basic Proportionality Theorem (BPT), we have:
\( \frac{OA}{AP} = \frac{OB}{BQ} \) .........(1)
2. Consider \( \triangle POR \). Since \( AC \parallel PR \) (given), by the Basic Proportionality Theorem (BPT), we have:
\( \frac{OA}{AP} = \frac{OC}{CR} \) .........(2)
3. From equation (1) and equation (2), since both ratios are equal to \( \frac{OA}{AP} \), we can equate them:
\( \frac{OB}{BQ} = \frac{OC}{CR} \)
4. Now, consider \( \triangle OQR \). We have shown that the line BC divides the sides OQ and OR in the same ratio. Therefore, by the converse of the Basic Proportionality Theorem (BPT), BC || QR.
Hence, proved.
In simple words: We used the BPT theorem in two smaller triangles, OPQ and OPR, because we knew about two pairs of parallel lines. This gave us two equations with equal ratios. Since both equations shared a common ratio involving OA and AP, we could link the other parts together, showing that \( \frac{OB}{BQ} \) is equal to \( \frac{OC}{CR} \). Then, by using the reverse rule of BPT in the big triangle OQR, we could confirm that BC is also parallel to QR.

Exam Tip: For theorems involving three parallel lines or connecting multiple triangles, identify the common point (O in this case) and apply BPT step-by-step to each relevant triangle. The key is finding a shared ratio to connect the different parts of the figure.

Question 7. Using theorem 6.1 (i.e. Basic proportionality theorem), prove that line drawn through the mid point of one side of a triangle parallel to another side bisects the third side (Recall that you have proved it in class IX) (CBSE 2012)

Answer:
Given:
In \( \triangle ABC \), D is the midpoint of side AB.
A line segment DE is drawn through D such that \( DE \parallel BC \).
To prove:
E is the midpoint of AC (i.e., \( AE = EC \)).
Proof:
1. In \( \triangle ABC \), since \( DE \parallel BC \) (given), by the Basic Proportionality Theorem (BPT), we have:
\( \frac{AD}{DB} = \frac{AE}{EC} \) .........(1)
2. We are given that D is the midpoint of AB. This means that AD = DB.
Therefore, \( \frac{AD}{DB} = 1 \).
3. Substitute \( \frac{AD}{DB} = 1 \) into equation (1):
\( 1 = \frac{AE}{EC} \)
4. Multiplying both sides by EC, we get:
\( EC = AE \)
Thus, E is the midpoint of AC.
Hence, proved.
In simple words: We used the BPT because we had a line parallel to one side of a triangle, starting from the midpoint of another side. This allowed us to set up a ratio. Since the starting point was a midpoint, the first ratio became 1. Setting this equal to the second ratio showed that the line also split the third side into two equal parts, meaning it also passed through its midpoint.

Exam Tip: This is the Midpoint Theorem, which is a direct consequence of BPT. When proving it, clearly state that if a point is a midpoint, the ratio of the segments it creates is 1. This simplifies the BPT equation and leads directly to the conclusion.

Question 8. Using theorem 6.2 (i.e. Converse of basic proportionality theorem), prove that the line joining the midpoints of any two sides a triangle is a parallel to the third side (Recall that you have done it in class IX)

Answer:
Given:
In \( \triangle ABC \), D is the midpoint of side AB, and E is the midpoint of side AC.
DE is the line segment joining D and E.
To prove:
\( DE \parallel BC \)
Proof:
1. Since D is the midpoint of AB, we have \( AD = DB \).
Therefore, \( \frac{AD}{DB} = 1 \) .........(1)
2. Since E is the midpoint of AC, we have \( AE = EC \).
Therefore, \( \frac{AE}{EC} = 1 \) .........(2)
3. From equation (1) and equation (2), we can conclude:
\( \frac{AD}{DB} = \frac{AE}{EC} \)
4. Now, consider \( \triangle ABC \). Since the line segment DE divides sides AB and AC in the same ratio, by the converse of the Basic Proportionality Theorem (BPT), DE || BC.
Hence, proved.
In simple words: We started with D and E being the midpoints of two sides of the triangle. This meant that the ratio of the parts on each side was 1. Since these ratios were equal, we could use the reverse rule of BPT. This rule states that if a line divides two sides of a triangle proportionally, then that line must be parallel to the third side. So, we proved DE is parallel to BC.

Exam Tip: This is the converse of the Midpoint Theorem. The proof relies on using the definition of a midpoint to establish that the ratios \( \frac{AD}{DB} \) and \( \frac{AE}{EC} \) are both equal to 1, and then applying the converse of BPT.

Question 9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \( \frac{AO}{BO} = \frac{CO}{DO} \). (CBSE 2004)

Answer:
Given:
ABCD is a trapezium with \( AB \parallel DC \). The diagonals AC and BD intersect at O.
To show:
\( \frac{AO}{BO} = \frac{CO}{DO} \)
Construction:
Draw a line OE through O parallel to AB, intersecting AD at E.
Proof:
1. Consider \( \triangle DAB \). Since \( OE \parallel AB \) (by construction), by the Basic Proportionality Theorem (BPT), we have:
\( \frac{DE}{EA} = \frac{DO}{OB} \) .........(1)
2. Now, consider \( \triangle ADC \). We know that \( AB \parallel DC \) (given) and \( OE \parallel AB \) (by construction). Therefore, \( OE \parallel DC \).
3. In \( \triangle ADC \), since \( OE \parallel DC \), by the Basic Proportionality Theorem (BPT), we have:
\( \frac{AE}{ED} = \frac{AO}{OC} \) .........(2)
4. From equation (1), we have \( \frac{DO}{OB} = \frac{DE}{EA} \). Inverting this, we get \( \frac{OB}{DO} = \frac{EA}{DE} \).
From equation (2), we have \( \frac{AO}{OC} = \frac{AE}{ED} \).
5. We can rewrite equation (1) as \( \frac{DO}{BO} = \frac{DE}{AE} \).
We can rewrite equation (2) as \( \frac{AO}{CO} = \frac{AE}{DE} \).
From (1) and (2), we notice that \( \frac{DE}{AE} \) is the reciprocal of \( \frac{AE}{DE} \). So, \( \frac{DO}{BO} \) is the reciprocal of \( \frac{AO}{CO} \).
This means \( \frac{DO}{BO} = \frac{CO}{AO} \).
6. By Alternendo (swapping the means), we get:
\( \frac{AO}{BO} = \frac{CO}{DO} \)
Hence, proved.
In simple words: We drew an extra line through O that was parallel to AB. This line also became parallel to DC because AB and DC were already parallel. Then, we used BPT in two different triangles (DAB and ADC). This gave us ratios of sides involving the segments created by the new line. By carefully comparing and rearranging these ratios, we found that \( \frac{AO}{BO} \) is equal to \( \frac{CO}{DO} \), which is what we needed to show.

Exam Tip: For problems involving trapeziums and intersecting diagonals, the common strategy is to draw a line through the intersection point parallel to the parallel sides. This creates two triangles where BPT can be applied, linking the ratios of the diagonal segments.

Question 10. The diagonals of quadrilateral ABCD intersect each other at the point O such that \( \frac{AO}{BO} = \frac{CO}{DO} \), show that ABCD is a trapezium. (CBSE 2012)

Answer:
Given:
ABCD is a quadrilateral where diagonals AC and BD intersect at O, such that \( \frac{AO}{BO} = \frac{CO}{DO} \).
To prove:
ABCD is a trapezium (i.e., \( AB \parallel DC \)).
Construction:
Draw a line OE through O such that \( OE \parallel AB \), where E is a point on AD.
Proof:
1. Consider \( \triangle DAB \). Since \( OE \parallel AB \) (by construction), by the Basic Proportionality Theorem (BPT), we have:
\( \frac{DE}{AE} = \frac{DO}{BO} \) .........(1)
2. We are given that \( \frac{AO}{BO} = \frac{CO}{DO} \).
Rearranging this by Alternendo, we get \( \frac{AO}{CO} = \frac{BO}{DO} \).
By Invertendo on this, we get \( \frac{CO}{AO} = \frac{DO}{BO} \) .........(2)
3. From equation (1) and equation (2), since both ratios are equal to \( \frac{DO}{BO} \), we can equate the other parts:
\( \frac{DE}{AE} = \frac{CO}{AO} \)
4. Now, consider \( \triangle ADC \). In this triangle, we have \( \frac{AE}{DE} = \frac{AO}{CO} \).
Since the line EO divides the sides AD and AC in the same ratio, by the converse of the Basic Proportionality Theorem (BPT), \( OE \parallel DC \).
5. We constructed OE such that \( OE \parallel AB \).
We have now proved that \( OE \parallel DC \).
Since \( OE \parallel AB \) and \( OE \parallel DC \), it follows that \( AB \parallel DC \).
6. A quadrilateral with one pair of parallel sides is a trapezium.
Therefore, ABCD is a trapezium.
Hence, proved.
In simple words: We were given a proportional relationship between the segments of the diagonals. To prove it's a trapezium, we need to show that two sides are parallel. We drew an auxiliary line through the intersection point parallel to one of the potential parallel sides. By using BPT and the given ratio, we showed that this auxiliary line was also parallel to the other potential parallel side. Since the auxiliary line was parallel to both, the two main sides must be parallel to each other, making the figure a trapezium.

Exam Tip: This is the converse of the trapezium property proof. The key is to introduce a construction line (OE parallel to AB) and then use the given diagonal ratio with BPT to establish that this construction line is also parallel to the other side (DC), thus proving the original sides are parallel.