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 and prove the Basic Proportionality Theorem (Thales' theorem): If a line parallel to one side of a triangle intersects the other two sides, it divides them proportionally. Apply this to $△ A BC$ with $L M ∥ CB$ and $△ A N D$ with $L N ∥ C D$ to show $\frac{A M}{MB} = \frac{A N}{N D}$ .

Step-by-step working

1
State BPT: Line parallel to one side divides other sides in same ratio
1 mark
2
Prove BPT using similar triangles or area method
1 mark
3
Apply BPT to $△ A BC$ with $L M ∥ CB$ : $\frac{A M}{MB} = \frac{A L}{L C}$
1 mark
4
Apply BPT to $△ A N D$ with $L N ∥ C D$ : $\frac{A N}{N D} = \frac{A L}{L C}$
1 mark
5
Conclude $\frac{A M}{MB} = \frac{A N}{N D}$
1 mark

Concepts used

basic proportionality theoremthales theoremparallel linessimilar triangles

Approach

Step-by-step working

Concepts used

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