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Triangles
EXERCISE 6.2
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
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).
Step-by-Step Solution
Key Concept: Apply the converse of the Mid‑point theorem (Theorem 6.2). If a line through the midpoint of one side of a triangle is parallel to a second side, then it bisects the third side. By constructing such a line and using the uniqueness of the midpoint, we obtain the required parallelism.
1. Consider triangle \(\triangle ABC\).
Let \(D\) be the midpoint of side \(AB\) and \(E\) be the midpoint of side \(AC\).
2. Draw a line through \(D\) parallel to \(BC\).
Let this line intersect side \(AC\) at a point \(F\). Thus, \(DF \parallel BC\).
3. Apply Theorem 6.2 (converse of the Mid‑point theorem).
Since \(D\) is the midpoint of \(AB\) and \(DF\) is drawn through \(D\) parallel to \(BC\), Theorem 6.2 tells us that \(DF\) bisects the third side \(AC\). Hence, \(F\) is the midpoint of \(AC\).
4. Identify the midpoint on \(AC\).
By definition, the midpoint of \(AC\) is the point \(E\). Because a side has only one midpoint, we must have \(F \equiv E\).
5. Conclude the required parallelism.
The line joining the midpoints \(D\) and \(E\) is exactly the line \(DF\) constructed in step 2. Therefore, \(DE \parallel BC\).
6. Result.
Hence, the line joining the mid‑points of any two sides of a triangle is parallel to the third side.
Thus, \(DE \parallel BC\).
Let \(D\) be the midpoint of side \(AB\) and \(E\) be the midpoint of side \(AC\).
2. Draw a line through \(D\) parallel to \(BC\).
Let this line intersect side \(AC\) at a point \(F\). Thus, \(DF \parallel BC\).
3. Apply Theorem 6.2 (converse of the Mid‑point theorem).
Since \(D\) is the midpoint of \(AB\) and \(DF\) is drawn through \(D\) parallel to \(BC\), Theorem 6.2 tells us that \(DF\) bisects the third side \(AC\). Hence, \(F\) is the midpoint of \(AC\).
4. Identify the midpoint on \(AC\).
By definition, the midpoint of \(AC\) is the point \(E\). Because a side has only one midpoint, we must have \(F \equiv E\).
5. Conclude the required parallelism.
The line joining the midpoints \(D\) and \(E\) is exactly the line \(DF\) constructed in step 2. Therefore, \(DE \parallel BC\).
6. Result.
Hence, the line joining the mid‑points of any two sides of a triangle is parallel to the third side.
Thus, \(DE \parallel BC\).
Correct Answer: The line joining the mid‑points of any two sides of a triangle is parallel to the third side; i.e., \(DE \parallel BC\).
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