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In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC) - Mathematics

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प्रश्न

In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)

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उत्तर

Given: In ΔABC, L and M are points on AB and AC respectively such that LM || BC.

To prove: ar (ΔLOB) = ar (ΔMOC)

Proof: We know that, triangles on the same base and between the same base between the same parallels are equal in area.

Hence, ΔLBC and ΔMBC lie on the same base BC and between the same parallels BC and LM.

So, ar (ΔLBC) = ar (ΔMBC)

⇒ ar (ΔLOB) + ar (ΔBOC) = ar (ΔMOC) + ar (ΔBOC)

On eliminating D ar (ΔBOC) from both sides, we get

ar (ΔLOB) = ar (ΔMOC)

Hence proved.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Areas of Parallelograms & Triangles - Exercise 9.4 [पृष्ठ ९५]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 9
अध्याय 9 Areas of Parallelograms & Triangles
Exercise 9.4 | Q 6. | पृष्ठ ९५

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