Medians

Geometry Level 2

In △ A B C , \triangle ABC, AD is the median to BC. E is the midpoint of AD. F is the midpoint of AE. What fraction of a r ( A B C ) ar(ABC ) is a r ( A F B ) ar(AFB) .

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1 10 \frac{1}{10} 1 4 \frac{1}{4} 1 8 \frac{1}{8} 1 6 \frac{1}{6}

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Vignesh Rao by Vignesh Rao , 20, India

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1 solution

Akhil Bansal
Dec 24, 2015

⇒ a r ( A B C ) = 2 a r ( A B D ) ( AD is median ) \Rightarrow ar(\color{#3D99F6}{ABC}) = 2ar(\color{#D61F06}{ABD}) \quad \quad \quad \quad (\text{AD is median})
⇒ a r ( A B D ) = 2 a r ( A B E ) ( BE is median ) \Rightarrow ar(\color{#D61F06}{ABD}) = 2ar(\color{#69047E}{ABE}) \quad \quad \quad \quad (\text{BE is median})
⇒ a r ( A B E ) = 2 a r ( A B F ) ( BF is median ) \Rightarrow ar(\color{#69047E}{ABE}) = 2ar(\color{#20A900}{ABF}) \quad \quad \quad \quad (\text{BF is median})

Then, ∴ a r ( A B F ) a r ( A B C ) = 1 8 \large \therefore \dfrac{ar(\color{#20A900}{ABF})}{ar(\color{#3D99F6}{ABC})} = \dfrac{1}{8}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice Solution.

Vignesh Rao - 5 years, 5 months ago

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