Triangles Ex 6.2 Class 10 NCERT: BPT and its Converse

Key Theorems for Exercise 6.2

Theorem 6.1: Basic Proportionality Theorem (BPT) or Thales' Theorem

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

Theorem 6.2: Converse of the Basic Proportionality Theorem

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

Understanding the Basic Proportionality Theorem (BPT)

The Basic Proportionality Theorem is a cornerstone of this exercise. Imagine a triangle, let's call it ΔABC. Now, draw a line DE such that D is on side AB and E is on side AC. If this line DE is parallel to the third side, BC (i.e., DE || BC), then the BPT guarantees a special relationship.

The theorem states that the line DE divides the sides AB and AC in the same ratio. Mathematically, this means:

AD / DB = AE / EC

Think of it like a balanced scale. The ratio of the parts on one side (AB) is exactly equal to the ratio of the parts on the other side (AC). This simple equation is incredibly useful for finding the length of an unknown segment. The Converse of BPT works in reverse. If you can prove that AD/DB = AE/EC, then you can conclude that the line DE must be parallel to BC. This is your tool for proving lines are parallel.

Worked Example: Applying the BPT

• Problem: In ΔABC, a line DE is parallel to BC. If AD = 1.5 cm, DB = 3 cm, and AE = 1 cm, find the length of EC. Step 1: Identify the Given Information and the Theorem to Use. We are given a triangle ΔABC where DE || BC. The lengths of three segments are provided (AD, DB, AE), and we need to find the fourth (EC). Since we know the line is parallel, we must use the Basic Proportionality Theorem (BPT). Step 2: State the BPT Ratio. According to BPT, if DE || BC, then: AD / DB = AE / EC Step 3: Substitute the Known Values into the Ratio. Substitute the given lengths into the equation: 1.5 / 3 = 1 / EC Step 4: Solve for the Unknown Variable (EC). To solve for EC, we can cross-multiply: 1.5 EC = 3 1 1.5 * EC = 3 EC = 3 / 1.5 EC = 2 cm Final Answer: The length of EC is 2 cm.

Exam Tips & Common Mistakes

Many students make small errors with BPT that can cost them marks. Here's what to watch out for:

1. Confusing the Ratios: The standard BPT ratio is AD/DB = AE/EC. A common mistake is to write AD/AB = AE/EC. While a corollary of BPT states that AD/AB = AE/AC = DE/BC, it's different from the basic theorem. Always check which parts of the side you are comparing. Stick to the basic part1/part2 ratio unless the question requires comparing a part to the whole side.

1. Applying BPT without Parallel Lines: You can only use BPT (AD/DB = AE/EC) if you are given that the lines are parallel (DE || BC). If you need to prove lines are parallel, you must first show the ratio is equal and then state that by the Converse of BPT, the lines are parallel. Don't mix them up!

1. Calculation Errors: The problems often involve decimals or fractions. Be very careful with your cross-multiplication and division. Double-check your arithmetic before writing the final answer.

Practice Questions with Solutions

• Q: In ΔPQR, E and F are points on the sides PQ and PR respectively. If PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm, determine whether EF is parallel to QR. A: Step 1: To check if EF || QR, we must use the Converse of the Basic Proportionality Theorem. We need to verify if PE/QE = PF/RF. Step 2: Calculate the ratio of the sides. Ratio on side PQ: PE/QE = 4 / 4.5. To simplify, multiply numerator and denominator by 10: 40/45 = 8/9. Ratio on side PR: PF/RF = 8 / 9. Step 3: Compare the ratios. Since PE/QE = 8/9 and PF/RF = 8/9, the ratios are equal. Final answer: Because the line EF divides the sides PQ and PR in the same ratio, by the Converse of the Basic Proportionality Theorem, EF is parallel to QR.
• Q: In a triangle ABC, D and E are points on sides AB and AC respectively such that DE || BC. If AD = x, DB = x - 2, AE = x + 2 and EC = x - 1, find the value of x. A: Step 1: Since it is given that DE || BC, we can apply the Basic Proportionality Theorem (BPT). Step 2: Set up the equation according to BPT: AD/DB = AE/EC. Substitute the given expressions: x / (x - 2) = (x + 2) / (x - 1). Step 3: Cross-multiply to solve for x. x (x - 1) = (x + 2) (x - 2) x² - x = x² - 4 (using the identity (a+b)(a-b) = a² - b²). Step 4: Simplify the equation. -x = -4 x = 4. Final answer: The value of x is 4.
• Q: In the given figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD. (Figure shows a triangle ABC with point M on AB. A point D is outside the triangle such that A, C, D form another triangle, and N is on AD. A is a common vertex.) A: Step 1: Consider ΔABC. We are given that LM || CB. By applying the Basic Proportionality Theorem, we get: AM/MB = AL/LC --- (Equation 1) However, using a corollary of BPT, we can also write: AM/AB = AL/AC --- (Equation i) Step 2: Now, consider ΔADC. We are given that LN || CD. By applying the BPT corollary, we get: AN/AD = AL/AC --- (Equation ii) Step 3: Compare Equation (i) and Equation (ii). Both equations have a common right-hand side (RHS), which is AL/AC. Step 4: Since the RHS of both equations are equal, their left-hand sides (LHS) must also be equal. Therefore, AM/AB = AN/AD. Final answer: Hence, proved.
• Q: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB. A: Step 1: Identify properties of a parallelogram. In parallelogram ABCD, AB || DC (and hence AB || FC) and AD || BC (and hence AE || BC). Step 2: Consider ΔABE and ΔCFB. Since AE || BC and BE is a transversal, we have ∠AEB = ∠CBF (Alternate Interior Angles). Step 3: Since AB || FC and BE is a transversal, we have ∠ABE = ∠BFC (Alternate Interior Angles). Also, in a parallelogram opposite angles are equal, so ∠DAB = ∠BCD. The angle at A in ΔABE is ∠EAB. The angle at C in ΔCFB is ∠BCF. ∠EAB = ∠BCF because they are opposite angles of the parallelogram. Step 4: We can also use Vertically Opposite Angles. However, let's use the property that opposite angles of a parallelogram are equal. ∠A (in ΔABE) = ∠C (in ΔCFB). So, ∠BAE = ∠FCB. Step 5: Since two angles of ΔABE are equal to two corresponding angles of ΔCFB (∠AEB = ∠CBF and ∠BAE = ∠FCB), the two triangles are similar by AA similarity criterion. Final answer: By AA similarity criterion, ΔABE ~ ΔCFB. (Note: While this question is often in the same chapter, it uses similarity criteria more directly than BPT, which is a good link between the concepts).

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