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In the adjoining figure, AD and BE are medians of △ABC. If DF || BE, prove that CF = 14AC.\dfrac{1}{4}AC.

In the adjoining figure, AD and BE are medians of △ABC. If DF || BE, prove that CF = (1/4)AC. Mid-point Theorem, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Mid-point Theorem

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Answer

In △BCE,

D is the midpoint of BC (As AD is median)

DF || BE

∴ F is the midpoint of CE (By converse of mid-point theorem).

⇒ CF = 12CE\dfrac{1}{2}CE …….(i)

Given,

BE is median

∴ CE = 12AC\dfrac{1}{2}AC

Substituting value of CE in (i) we get,

⇒CF=12CE⇒CF=12×12AC⇒CF=14AC.\Rightarrow CF = \dfrac{1}{2}CE \\[1em] \Rightarrow CF = \dfrac{1}{2} \times \dfrac{1}{2}AC \\[1em] \Rightarrow CF = \dfrac{1}{4}AC.

Hence, proved that CF=14AC.CF = \dfrac{1}{4}AC.

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