Question 5 of 14intermediate🔧 ApplyLong Answer3 marks

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. (Recall that you have done it in Class IX).

Correct Answer

Given: A triangle ABCABC in which DD and EE are the mid-points of sides ABAB and ACAC respectively.

To Prove: DE∥BCDE \parallel BC.

Proof:

Since DD is the mid-point of ABAB,

AD=DBAD = DB

Therefore,

ADDB=1…(1)\frac{AD}{DB} = 1 \quad \ldots \text{(1)}

Since EE is the mid-point of ACAC,

AE=ECAE = EC

Therefore,

AEEC=1…(2)\frac{AE}{EC} = 1 \quad \ldots \text{(2)}

From equations (1) and (2), we get:

ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}

Thus, the line DEDE divides the two sides ABAB and ACAC of triangle ABCABC in the same ratio.

By Theorem 6.2: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Therefore, DE∥BCDE \parallel BC.

Hence proved.

Exercise: EXERCISE 6.2 | Q: 8 | (Page Page 13)
For More Understanding

Explanation

This question directly applies Theorem 6.2 (converse of Basic Proportionality Theorem). The key insight is recognizing that when DD and EE are midpoints, the ratios ADDB\frac{AD}{DB} and AEEC\frac{AE}{EC} both equal 11.

Since the line DEDE divides both sides ABAB and ACAC in the same ratio (1:11:1), by Theorem 6.2, DEDE must be parallel to the third side BCBC. This is a fundamental result known as the Mid-point Theorem.

Solution Steps

  1. Step 1: Identify that DD is midpoint of ABAB, so AD=DBAD = DB, giving ADDB=1\frac{AD}{DB} = 1

  2. Step 2: Identify that EE is midpoint of ACAC, so AE=ECAE = EC, giving AEEC=1\frac{AE}{EC} = 1

  3. Step 3: Conclude that ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC} (both equal to 11)

  4. Step 4: Apply Theorem 6.2: Since DEDE divides ABAB and ACAC in the same ratio, DE∥BCDE \parallel BC

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