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The median AD of a triangle $\mathrm{ABC}$ is bisected at $\mathrm{F}$, and $\mathrm{BF}$ is produced to meet the side $A C$ in $P$. If $A P=\lambda A C$, then what is the value of $\lambda$ ?
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The correct answer is:
$\frac{1}{3}$
Given, $\mathrm{AD}$ is the median of $\triangle \mathrm{ABC} . \mathrm{F}$ is mid-point of AD and $\mathrm{BF}$ is produced to meet the side $\mathrm{AC}$ in $\mathrm{P}$.
Draw DQ $\| \mathrm{FP}$ In $\Delta \mathrm{ADQ}, \mathrm{F}$ is mid-point of $\mathrm{AD}$ and $\mathrm{FP} \| \mathrm{DQ}$. Pis mid-point of $A Q$ (converse of mid-point theorem) $\Rightarrow \mathrm{AP}=\mathrm{PQ}$ ...(1)
In $\Delta \mathrm{BCP}, \mathrm{D}$ is mid-point of $\mathrm{BC}$ and $\mathrm{DQ} \| \mathrm{BP}$.
$\therefore$ Q is midpoint of $P C$. $\Rightarrow \mathrm{PQ}=\mathrm{QC}$ ...(2)
From $(1),(2)$ we get, $A P=P Q=Q C$ From figure, $\mathrm{AP}+\mathrm{PQ}+\mathrm{QC}=\mathrm{AC}$
$\Rightarrow \mathrm{AP}+\mathrm{AP}+\mathrm{AP}=\mathrm{AC}$
$\Rightarrow 3 \times \mathrm{AP}=\mathrm{AC}$
$\Rightarrow \mathrm{AP}=\frac{1}{3} \times \mathrm{AC} \quad \therefore \quad \lambda=\frac{1}{3}$
Draw DQ $\| \mathrm{FP}$ In $\Delta \mathrm{ADQ}, \mathrm{F}$ is mid-point of $\mathrm{AD}$ and $\mathrm{FP} \| \mathrm{DQ}$. Pis mid-point of $A Q$ (converse of mid-point theorem) $\Rightarrow \mathrm{AP}=\mathrm{PQ}$ ...(1)
In $\Delta \mathrm{BCP}, \mathrm{D}$ is mid-point of $\mathrm{BC}$ and $\mathrm{DQ} \| \mathrm{BP}$.
$\therefore$ Q is midpoint of $P C$. $\Rightarrow \mathrm{PQ}=\mathrm{QC}$ ...(2)
From $(1),(2)$ we get, $A P=P Q=Q C$ From figure, $\mathrm{AP}+\mathrm{PQ}+\mathrm{QC}=\mathrm{AC}$
$\Rightarrow \mathrm{AP}+\mathrm{AP}+\mathrm{AP}=\mathrm{AC}$
$\Rightarrow 3 \times \mathrm{AP}=\mathrm{AC}$
$\Rightarrow \mathrm{AP}=\frac{1}{3} \times \mathrm{AC} \quad \therefore \quad \lambda=\frac{1}{3}$
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