Question

(a) 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.

Approach

Prove Basic Proportionality Theorem (Thales' theorem): 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. Standard proof: construct figure, use properties of similar triangles formed by the parallel line, and establish the ratio equality.

Step-by-step working

1
Draw $△ A BC$ and line $D E ∥ BC$ intersecting $A B$ at $D$ and $A C$ at $E$ .
1 mark
2
State: To prove $\frac{A D}{D B} = \frac{A E}{EC}$ . Construction: Join $BE$ , $C D$ ; draw $D M ⊥ A C$ and $EN ⊥ A B$ .
1 mark
3
Proof: Area( $△ A D E$ ) / Area( $△ D BE$ ) $= \frac{\frac{1}{2} \cdot A D \cdot EN}{\frac{1}{2} \cdot D B \cdot EN} = \frac{A D}{D B}$ . Similarly, Area( $△ A D E$ ) / Area( $△ D EC$ ) $= \frac{A E}{EC}$ . Since $D E ∥ BC$ , $△ D BE$ and $△ D EC$ have the same base $D E$ and equal heights, so Area( $△ D BE$ ) = Area( $△ D EC$ ). Hence $\frac{A D}{D B} = \frac{A E}{EC}$ .
3 marks

Concepts used

basic proportionality theoremproveparallel linestrianglethales theorem

Approach

Step-by-step working

Concepts used

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