Question 4 of 14intermediate🔧 ApplyLong Answer3 marks

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

Correct Answer

Given: In triangle ABC, D is the mid-point of side AB. A line DE is drawn through D parallel to BC, intersecting AC at E.

To Prove: E is the mid-point of AC, i.e., AE=ECAE = EC.

Proof:

In triangle ABC, DE∥BCDE \parallel BC (Given). By Theorem 6.1, 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.

Therefore,

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

Since D is the mid-point of AB, AD=DBAD = DB, which gives

ADDB=1\frac{AD}{DB} = 1

Substituting this value:

1=AEEC1 = \frac{AE}{EC}

which implies AE=ECAE = EC.

Hence, E bisects side AC. Therefore, the line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

∴ E is the mid-point of AC, i.e., AE=ECAE = EC.

Exercise: EXERCISE 6.2 | Q: 7 | (Page Page 13)
For More Understanding

Explanation

This proof applies Theorem 6.1 (Basic Proportionality Theorem) to establish the converse of the Mid-Point Theorem. The key insight is that when D is a mid-point, the ratio ADDB\frac{AD}{DB} equals 1, and by Theorem 6.1, this ratio must equal AEEC\frac{AE}{EC}, forcing E to also be a mid-point.

Solution Steps

  1. Step 1: Identify that DE∥BCDE \parallel BC in triangle ABC, with D on AB and E on AC.

  2. Step 2: Apply Theorem 6.1 to get ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}.

  3. Step 3: Use the fact that D is mid-point of AB, so AD=DBAD = DB and ADDB=1\frac{AD}{DB} = 1.

  4. Step 4: Conclude that AEEC=1\frac{AE}{EC} = 1, hence AE=ECAE = EC, proving E is the mid-point of AC.

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