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NCERT Class 10 Mathematics Exercise 6.2 Solutions: Triangles

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of Exercise 6.2 in NCERT's Class 10 Mathematics book, chapter “Triangles”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

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How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.

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  1. Ex 6.2 Q1

    DE ∥ BC in both triangles of Fig. 6.17. (i) AD = 1.5 cm, DB = 3 cm, AE = 1 cm: find EC. (ii) DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm: find AD.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. (i) Since DE ∥ BC, by the Basic Proportionality Theorem, AD/DB = AE/EC.
      why?

      A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem); its converse also holds.

    2. (i) Substitute the given values: 1.5/3 = 1/EC.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    3. (i) Simplify the left side: 0.5 = 1/EC.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    4. (i) Solve for EC: EC = 1/0.5 = 2 cm.
    5. (ii) Since DE ∥ BC, by the Basic Proportionality Theorem, AD/DB = AE/EC.
      why?

      A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem); its converse also holds.

    6. (ii) Substitute the given values: AD/7.2 = 1.8/5.4.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    7. (ii) Simplify the right side: 1.8/5.4 = 1/3.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    8. (ii) Solve for AD: AD = 7.2 × (1/3) = 2.4 cm.

    Answer: (i) EC = 2 cm; (ii) AD = 2.4 cm

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  2. Ex 6.2 Q2

    E on PQ and F on PR of △PQR. Is EF ∥ QR when (i) PE = 3.9, EQ = 3, PF = 3.6, FR = 2.4 (ii) PE = 4, QE = 4.5, PF = 8, RF = 9 (iii) PQ = 1.28, PR = 2.56, PE = 0.18, PF = 0.36 (all in cm)?

    1. (i) Calculate the ratio PE/EQ using the given values: 3.9/3 = 1.3.
    2. (i) Calculate the ratio PF/FR using the given values: 3.6/2.4 = 1.5.
    3. (i) Compare the ratios: since 1.3 ≠ 1.5, the segments are not divided in the same ratio.
    4. (i) Conclude that EF is not parallel to QR.
    5. (ii) Calculate the ratio PE/EQ using the given values: 4/4.5 = 8/9.
    6. (ii) Calculate the ratio PF/FR using the given values: 8/9.
    7. (ii) Compare the ratios: since 8/9 = 8/9, the segments are divided in the same ratio.
    8. (ii) Conclude that EF is parallel to QR.
    9. (iii) Calculate the length of EQ by subtracting PE from PQ: 1.28 − 0.18 = 1.10 cm.
    10. (iii) Calculate the length of FR by subtracting PF from PR: 2.56 − 0.36 = 2.20 cm.
    11. (iii) Calculate the ratio PE/EQ: 0.18/1.10 = 18/110 = 9/55.
    12. (iii) Calculate the ratio PF/FR: 0.36/2.20 = 36/220 = 9/55.
    13. (iii) Compare the ratios: since 9/55 = 9/55, the segments are divided in the same ratio.
    14. (iii) Conclude that EF is parallel to QR.

    Answer: (i) EF not parallel to QR; (ii) EF ∥ QR; (iii) EF ∥ QR

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  3. Ex 6.2 Q3

    In Fig. 6.18, LM ∥ CB and LN ∥ CD. Prove AM/AB = AN/AD.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. In △ABC, since LM ∥ CB, by the Basic Proportionality Theorem (Thales' Theorem), AM/MB = AL/LC.
      why?

      A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem); its converse also holds.

    2. In △ACD, since LN ∥ CD, by the Basic Proportionality Theorem, AN/ND = AL/LC.
      why?

      A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem); its converse also holds.

    3. From the two equations above, we see that AM/MB = AN/ND.
    4. Using the property of equal ratios (if a/b = c/d, then a/(a+b) = c/(c+d)), we get AM/(AM + MB) = AN/(AN + ND).
    5. Since AM + MB = AB and AN + ND = AD, the equation becomes AM/AB = AN/AD.
    6. Hence, AM/AB = AN/AD is proved.

    Answer: AM/AB = AN/AD

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  4. Ex 6.2 Q4

    In Fig. 6.19, DE ∥ AC and DF ∥ AE. Prove BF/FE = BE/EC.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. D lies on AB and E on BC (Fig. 6.19).
    2. In △ABC, DE ∥ AC, so by the Basic Proportionality Theorem, BD/DA = BE/EC.
      why?

      A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem); its converse also holds.

    3. In △ABE, DF ∥ AE, so by the same theorem, BD/DA = BF/FE.
    4. Both ratios equal BD/DA, so BF/FE = BE/EC.

    Answer: BF/FE = BE/EC (both equal BD/DA).

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  5. Ex 6.2 Q5

    In Fig. 6.20, DE ∥ OQ and DF ∥ OR. Show that EF ∥ QR.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. D lies on OP, E on PQ and F on PR (Fig. 6.20).
    2. In △POQ, DE ∥ OQ, so by the Basic Proportionality Theorem, PE/EQ = PD/DO.
      why?

      A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem); its converse also holds.

    3. In △POR, DF ∥ OR, so PF/FR = PD/DO.
    4. Hence PE/EQ = PF/FR: in △PQR, the line EF divides PQ and PR in the same ratio.
    5. By the converse of the Basic Proportionality Theorem, EF ∥ QR.
      why?

      A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem); its converse also holds.

    Answer: EF ∥ QR.

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  6. Ex 6.2 Q6

    In Fig. 6.21, A, B, C lie on OP, OQ, OR with AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. In △OPQ, AB ∥ PQ, so by the Basic Proportionality Theorem, OA/AP = OB/BQ.
      why?

      A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem); its converse also holds.

    2. In △OPR, AC ∥ PR, so OA/AP = OC/CR.
    3. Hence OB/BQ = OC/CR: in △OQR, BC divides OQ and OR in the same ratio.
    4. By the converse of the Basic Proportionality Theorem, BC ∥ QR.
      why?

      A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem); its converse also holds.

    Answer: BC ∥ QR.

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  7. Ex 6.2 Q7

    Use the Basic Proportionality Theorem to prove that a line through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

    1. Let △ABC be a triangle where D is the midpoint of side AB.
    2. Draw a line through D parallel to side BC, intersecting side AC at point E.
    3. By the Basic Proportionality Theorem, since DE ∥ BC, the ratio AD/DB equals AE/EC.
      why?

      A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem); its converse also holds.

    4. Since D is the midpoint of AB, the length of AD is equal to the length of DB.
    5. Therefore, the ratio AD/DB is equal to 1.
    6. Substituting this value into the proportion gives AE/EC = 1.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    7. This implies that the length of AE is equal to the length of EC.
    8. Hence, E is the midpoint of side AC, proving that the line bisects the third side.

    Answer: The line through the midpoint of one side, parallel to another side, bisects the third side.

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  8. Ex 6.2 Q8

    Use the converse of the Basic Proportionality Theorem to prove that the line joining the mid-points of two sides of a triangle is parallel to the third side.

    1. Let ABC be a triangle where D is the midpoint of side AB and E is the midpoint of side AC.
    2. Since D is the midpoint of AB, the ratio AD/DB is equal to 1.
    3. Since E is the midpoint of AC, the ratio AE/EC is equal to 1.
    4. Therefore, the ratio AD/DB is equal to the ratio AE/EC.
    5. By the Converse of the Basic Proportionality Theorem, if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
      why?

      A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (Basic Proportionality Theorem); its converse also holds.

    6. Hence, the line segment DE is parallel to the side BC.

    Answer: DE ∥ BC

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  9. Ex 6.2 Q9

    ABCD is a trapezium with AB ∥ DC whose diagonals meet at O. Show that AO/BO = CO/DO.

    1. Consider triangles AOB and COD.
    2. Since AB is parallel to DC, the alternate interior angles are equal, so ∠OAB = ∠OCD and ∠OBA = ∠ODC.
    3. The vertically opposite angles at O are equal, so ∠AOB = ∠COD.
    4. By the AAA similarity criterion, △AOB is similar to △COD.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    5. Since the triangles are similar, the ratios of their corresponding sides are equal.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    6. Therefore, AO/CO = BO/DO.
    7. Rearranging this proportion gives AO/BO = CO/DO.

    Answer: AO/BO = CO/DO

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  10. Ex 6.2 Q10

    The diagonals of quadrilateral ABCD meet at O with AO/BO = CO/DO. Show that ABCD is a trapezium.

    1. Given that in quadrilateral ABCD, diagonals AC and BD intersect at O such that AO/BO = CO/DO.
    2. We need to show that ABCD is a trapezium, which means we must prove that one pair of opposite sides is parallel.
    3. Consider triangles AOB and COD.
    4. From the given AO/BO = CO/DO, rearranging gives AO/CO = BO/DO, and ∠AOB = ∠COD (vertically opposite angles).
    5. Therefore, triangle AOB is similar to triangle COD by the Side-Angle-Side (SAS) similarity criterion.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    6. Since the triangles are similar, their corresponding angles are equal, so angle OAB = angle OCD.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    7. These are alternate interior angles formed by the transversal AC intersecting lines AB and CD.
    8. Since the alternate interior angles are equal, AB is parallel to CD.
    9. A quadrilateral with at least one pair of parallel sides is a trapezium.
    10. Hence, ABCD is a trapezium.

    Answer: ABCD is a trapezium.

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