Question

Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points divides the other two sides in the same ratio. Hence, in the figure given below, prove that $\frac{A M}{MB} = \frac{A N}{N D}$ where $L M ∥ CB$ and $L N ∥ C D$ .

Approach

State the Basic Proportionality Theorem (BPT): if a line parallel to one side of a triangle intersects the other two sides, it divides them proportionally. In the given figure, LM $∥$ CB and LN $∥$ CD. Apply BPT to $△ A BC$ and $△ A D C$ to show $\frac{A M}{MB} = \frac{A N}{N D}$ .

Step-by-step working

1
Draw correct figure with triangle sides intersected by parallel lines LM and LN.
1.5 marks
2
In $△ A BC$ , LM $∥$ CB. By BPT, $\frac{A M}{MB} = \frac{A L}{L C}$ .
1 mark
3
In $△ A D C$ , LN $∥$ CD. By BPT, $\frac{A N}{N D} = \frac{A L}{L C}$ .
0.5 marks
4
From steps 2 and 3, $\frac{A M}{MB} = \frac{A L}{L C} = \frac{A N}{N D}$ .
0.5 marks
5
Hence $\frac{A M}{MB} = \frac{A N}{N D}$ .
1.5 marks

Concepts used

basic proportionality theoremproveparallel linesratio

Approach

Step-by-step working

Concepts used

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