Question

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then prove that the other two sides are divided in the same ratio.

Approach

Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio (Basic Proportionality Theorem). In $△ A BC$ , $D E ∥ BC$ . Draw perpendiculars from D and E. Using area ratios, $\frac{a r ( △ A D E )}{a r ( △ B D E )} = \frac{A D}{D B}$ and $\frac{a r ( △ A D E )}{a r ( △ CE D )} = \frac{A E}{EC}$ . Since triangles BDE and CED have equal areas (same base DE, between same parallels), $\frac{A D}{D B} = \frac{A E}{EC}$ .

Step-by-step working

1
Draw correct figure with $△ A BC$ , $D E ∥ BC$ , perpendiculars DG and EF
0.5 marks
2
$\frac{a r ( △ A D E )}{a r ( △ B D E )} = \frac{\frac{1}{2} \times A D \times EF}{\frac{1}{2} \times D B \times EF} = \frac{A D}{D B}$
1 mark
3
$\frac{a r ( △ A D E )}{a r ( △ CE D )} = \frac{\frac{1}{2} \times A E \times D G}{\frac{1}{2} \times EC \times D G} = \frac{A E}{EC}$
1 mark
4
$a r (△ B D E) = a r (△ CE D)$ (same base DE, between parallels DE and BC)
1 mark
5
From steps 2, 3, 4: $\frac{A D}{D B} = \frac{A E}{EC}$
1.5 marks

Concepts used

basic proportionality theoremsimilar trianglesmedianprove

Approach

Step-by-step working

Concepts used

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