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Mid-point and Its Converse [Including Intercept Theorem] — MATHEMATICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9MATHEMATICSMid-point and Its Converse [Including Intercept Theorem]3 Marks
Question
In $\triangle ABC, D$ and $E$ are points on side $\text{AB}$ such that $\text{AD} = \text{DE} = \text{EB}.$ Through $D$ and $E,$ lines are drawn parallel to $\text{BC}$ which meet side $\text{AC}$ at points $F$ and $G$ respectively. Through $F$ and $G,$ lines are drawn parallel to $\text{AB}$ which meets side $\text{BC}$ at points $M$ and $N$ respectively. Prove that: $\text{BM} =\text{ MN} = \text{NC}.$
✓

Answer

The figure is shown below

For triangle $AEG$
$D$ is the mid$-$point of $\text{AE}$ and $\text{DF}\|\text{EG}\|\text{ BC}$
Therefore $F$ is the midpoint of $\text{AG}$.
$\text{AF} = \text{GF} \ldots . .(1)$
Again $\text{DF} \| \text{EG} \| \text{BC}\| \text{DE} = \text{BE},$
$\therefore \text{GF} = \text{GC} .....(2)$
$(1), (2)$ we get $\text{AF}=\text{GF}=\text{GC}$
Similarly Since $\text{GN} \| \text{FM} \|\text{AB}$ and $ \text{AF} = \text{GF} ,$
$\therefore \text{BM}=\text{MN}=\text{NC}$
Hence proved

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