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Triangles
EXERCISE 6.2
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
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).
Step-by-Step Solution
Key Concept: Apply the Mid‑point Theorem (Theorem 6.1) to the triangle formed by the given side and the side parallel to it. By constructing similar triangles, the intercepted segment on the third side is shown to be equal to the other intercepted segment, proving it is a midpoint.
1. Given: Triangle \(\triangle ABC\). Let \(M\) be the mid‑point of side \(AB\). Through \(M\) draw a line \(l\) parallel to side \(AC\) meeting \(BC\) at \(D\).
2. Apply Theorem 6.1 (Mid‑point Theorem) to \(\triangle ABC\):
- Since \(M\) is the midpoint of \(AB\) and \(l\parallel AC\), the segment \(MD\) is the line joining the mid‑point of one side to a point on the third side.
- By Theorem 6.1, the line joining the mid‑point of a side to a point on the third side that is parallel to the second side will intersect the third side at its midpoint.
3. Construct similar triangles:
- Because \(MD \parallel AC\), \(\angle MDB = \angle ACB\) (alternate interior angles).
- Also, \(\angle MDB = \angle ACB\) and \(\angle MBD = \angle ABC\) (common angle at \(B\)).
- Hence \(\triangle MBD \sim \triangle ACB\) (AA similarity).
4. Correspondence of sides:
- From similarity, \(\frac{MB}{AB} = \frac{BD}{CB}\).
- But \(MB = \frac{AB}{2}\) because \(M\) is the midpoint of \(AB\).
- Substituting, \(\frac{\frac{AB}{2}}{AB} = \frac{BD}{CB}\) \(\Rightarrow \frac{1}{2} = \frac{BD}{CB}\).
- Therefore \(BD = \frac{CB}{2}\).
5. Conclusion:
- Since \(BD = \frac{CB}{2}\), point \(D\) divides \(BC\) into two equal parts; i.e., \(D\) is the midpoint of \(BC\).
- Hence the line drawn through the mid‑point of one side of a triangle parallel to another side bisects the third side.
6. Remark: This is exactly the converse of the Mid‑point Theorem proved in Class IX, confirming the result.
2. Apply Theorem 6.1 (Mid‑point Theorem) to \(\triangle ABC\):
- Since \(M\) is the midpoint of \(AB\) and \(l\parallel AC\), the segment \(MD\) is the line joining the mid‑point of one side to a point on the third side.
- By Theorem 6.1, the line joining the mid‑point of a side to a point on the third side that is parallel to the second side will intersect the third side at its midpoint.
3. Construct similar triangles:
- Because \(MD \parallel AC\), \(\angle MDB = \angle ACB\) (alternate interior angles).
- Also, \(\angle MDB = \angle ACB\) and \(\angle MBD = \angle ABC\) (common angle at \(B\)).
- Hence \(\triangle MBD \sim \triangle ACB\) (AA similarity).
4. Correspondence of sides:
- From similarity, \(\frac{MB}{AB} = \frac{BD}{CB}\).
- But \(MB = \frac{AB}{2}\) because \(M\) is the midpoint of \(AB\).
- Substituting, \(\frac{\frac{AB}{2}}{AB} = \frac{BD}{CB}\) \(\Rightarrow \frac{1}{2} = \frac{BD}{CB}\).
- Therefore \(BD = \frac{CB}{2}\).
5. Conclusion:
- Since \(BD = \frac{CB}{2}\), point \(D\) divides \(BC\) into two equal parts; i.e., \(D\) is the midpoint of \(BC\).
- Hence the line drawn through the mid‑point of one side of a triangle parallel to another side bisects the third side.
6. Remark: This is exactly the converse of the Mid‑point Theorem proved in Class IX, confirming the result.
Correct Answer: The line through the midpoint of one side of a triangle and parallel to another side meets the third side at its midpoint; thus it bisects the third side.
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