Question

(a) 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.

(b) It is given that sides AB and AC and median AD of $△ A BC$ are respectively proportional to sides PQ and PR and median PM of another $△ PQR$ . Show that $△ A BC \sim △ PQR$ .

Approach

Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. Use the Basic Proportionality Theorem by constructing perpendiculars from $D$ and $E$ to the other sides and showing that the areas of triangles $A D E$ and $B D E$ have the same ratio as $A D : D B$ , and similarly for $A D E$ and $C D E$ , leading to $\frac{A D}{D B} = \frac{A E}{EC}$ .

Step-by-step working

1
Draw correct figure with $D E ∥ BC$ in $△ A BC$ .
0.5 marks
2
Construct perpendiculars $D G ⊥ A C$ and $EF ⊥ A B$ . Compute $\frac{ar ( △ A D E )}{ar ( △ B D E )} = \frac{\frac{1}{2} \times A D \times EF}{\frac{1}{2} \times D B \times EF} = \frac{A D}{D B}$ .
1.5 marks
3
Similarly, $\frac{ar ( △ A D E )}{ar ( △ C D E )} = \frac{\frac{1}{2} \times A E \times D G}{\frac{1}{2} \times EC \times D G} = \frac{A E}{EC}$ .
1.5 marks
4
Since $△ B D E$ and $△ C D E$ have the same base $D E$ and are between the same parallels $D E$ and $BC$ , $ar (△ B D E) = ar (△ C D E)$ .
1 mark
5
From the area ratios, $\frac{A D}{D B} = \frac{A E}{EC}$ . Proved.
0.5 marks

Concepts used

similar trianglesbasic proportionality theoremproveshow that

Approach

Step-by-step working

Concepts used

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