10thmath chapter 6 triangle

Class 10 math 

NCERT Solutions

 Exercise 6.2

Proportionality Theorem (BPT)

Statement:

If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

Diagram:

Mathematical Representation:

If DE || BC, then:

AD/DB = AE/EC

Proof:

 * Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.

 * Area of Triangles:

   * Area of triangle ADE = (1/2) * AE * DM

   * Area of triangle BDE = (1/2) * BD * DM

   * Area of triangle CDE = (1/2) * CE * DM

   * Area of triangle ADE = (1/2) * AD * EN

 * Ratio of Areas:

   * Area(ADE) / Area(BDE) = AE/BD

   * Area(ADE) / Area(CDE) = AD/CE

 * Equating Ratios:

   Since triangles BDE and CDE are on the same base DE and between the same parallels BC and DE, their areas are equal.

   Therefore, AE/BD = AD/CE.

Hence, the theorem is proved.

Converse of the Basic

Proportionality Theorem (BPT)

Statement:

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Diagram:

Mathematical Representation:

If AD/DB = AE/EC, then DE || BC.

Proof:

 * Assume the contrary: Let's assume that DE is not parallel to BC.

 * Draw a line DF parallel to BC: Draw a line DF parallel to BC, intersecting AC at F.

 * Apply BPT: Since DF || BC, we can apply BPT:

   AD/DB = AF/FC

 * Compare ratios:

   We are given that AD/DB = AE/EC.

   From steps 3 and 4, we have AF/FC = AE/EC.

 * Conclusion:

   This implies that F and E coincide (as they divide AC in the same ratio).

   Therefore, DE must be parallel to BC.

Hence, the converse of BPT is proved.

 Exercise 6.2  Solution using the Basic Proportionality Theorem (BPT):

1. In Fig. 6.17 (i), find EC:

 * We know that DE || BC.

 * Using BPT, we can write:

   AD/DB = AE/EC

 * Substituting the given values:

   1.5/3 = 1/EC

 * Solving for EC:

   EC = (3 * 1) / 1.5 = 2 cm

In Fig. 6.17 (ii), find AD:

 * We know that DE || BC.

 * Using BPT, we can write:

   AD/DB = AE/EC

 * Substituting the given values:

   AD/7.2 = 1.8/5.4

 * Solving for AD:

   AD = (7.2 * 1.8) / 5.4 = 2.4 cm

2. State whether EF || QR in each case:

 * (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, and FR = 2.4 cm:

   * Check if PE/EQ = PF/FR

   * 3.9/3 = 3.6/2.4

   * Both sides are equal to 1.3, so EF || QR.

 * (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, and RF = 9 cm:

   * Check if PE/EQ = PF/FR

   * 4/4.5 = 8/9

   * Both sides are equal to 8/9, so EF || QR.

 * (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, and PF = 0.36 cm:

   * Check if PE/EQ = PF/FR

   * 0.18/(1.28 - 0.18) = 0.36/(2.56 - 0.36)

   * Both sides are equal to 0.15, so EF || QR.

3. Prove AM/AB = AN/AD:

 * We are given that LM || CB and LN || CD.

 * Using BPT on triangle ABC with LM || BC:

   AM/MB = AL/LC

 * Using BPT on triangle ADC with LN || CD:

   AN/ND = AL/LC

 * From the above two equations, we can say:

   AM/MB = AN/ND

 * Adding 1 to both sides:

   (AM + MB)/MB = (AN + ND)/ND

   AB/MB = AD/ND

 * Taking reciprocals:

   MB/AB = ND/AD

 * Subtracting 1 from both sides:

   (MB - AB)/AB = (ND - AD)/AD

   -AM/AB = -AN/AD

 * Multiplying both sides by -1:

   AM/AB = AN/AD

4. Prove BF/FE = BE/EC:

 * We are given that DE || AC and DF || AE.

 * Using BPT on triangle ABC with DE || AC:

   BF/FE = BD/DA

 * Using BPT on triangle ABE with DF || AE:

   BD/DA = BE/EC

 * From the above two equations, we can say:

   BF/FE = BE/EC

Here's how to solve the problems from Exercise 6.2:

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

 * We can use the converse of the Basic Proportionality Theorem.

 * Since DE || OQ, we have OD/DP = OE/EP.

 * Similarly, since DF || OR, we have OF/FR = OD/DP.

 * Combining the two equations, we get OE/EP = OF/FR.

 * This means that EF || QR by the converse of BPT.

6. In Fig. 6.21, AB || PQ and AC || PR. Show that BC || QR.

 * We can use the same approach as in problem 5.

 * Since AB || PQ, we have AO/OP = BO/OQ.

 * Similarly, since AC || PR, we have AO/OP = CO/OR.

 * Combining the two equations, we get BO/OQ = CO/OR.

 * This means that BC || QR by the converse of BPT.

7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

 * Let D be the midpoint of AB in triangle ABC.

 * Draw DE parallel to BC.

 * By BPT, we have AD/DB = AE/EC.

 * Since AD = DB (D is the midpoint), we have AE = EC.

 * Therefore, DE bisects AC.

8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

 * Let D and E be the midpoints of AB and AC respectively in triangle ABC.

 * Join DE.

 * By the converse of BPT, if AD/DB = AE/EC, then DE || BC.

 * Since D and E are midpoints, AD = DB and AE = EC.

 * Therefore, AD/DB = AE/EC, and DE || BC.

9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

 * Since AB || DC, we have AO/BO = CO/DO by BPT applied to triangles AOB and COD.

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

 * We can use the converse of BPT.

 * Since AO/BO = CO/DO, we have AO/CO = BO/DO.

 * This means that AB || CD by the converse of BPT.

 * Therefore, ABCD is a trapezium.