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.
OR
(b) AD and PS are respectively, the medians of $△ A BC$ and $△ PQR$ . If $△ A BC \sim △ PQR$ , then prove that
(i) $△ A D C \sim △ PSR$
(ii) $\frac{A D}{PS} = \frac{BC}{QR}$

Approach

Prove: If a line parallel to one side of a triangle intersects the other two sides in distinct points, then it divides the sides proportionally. State BPT correctly with Given, To Prove, Construction (draw perpendiculars from D and E to AB and AC), and Proof using area ratios of triangles DBE and DCE on same base and between same parallels. Conclude $\frac{A D}{D B} = \frac{A E}{EC}$ .

Step-by-step working

1
State: In $△ A BC$ , $D E ∥ BC \Rightarrow \frac{A D}{D B} = \frac{A E}{EC}$ . Given, To Prove, Construction, Figure (draw perpendiculars DM $⊥$ AC, EN $⊥$ AB, join BE, CD).
2 marks
2
Area ratios: $\frac{ar ( △ D BE )}{ar ( △ A D E )} = \frac{D B}{A D}$ (same height EN). Similarly $\frac{ar ( △ D CE )}{ar ( △ A D E )} = \frac{EC}{A E}$ (same height DM).
1.5 marks
3
Since DE $∥$ BC, triangles DBE and DCE have equal areas (same base DE, between same parallels). Equate ratios: $\frac{D B}{A D} = \frac{EC}{A E} \Rightarrow \frac{A D}{D B} = \frac{A E}{EC}$ .
1 mark
4
Hence proved.
0.5 marks

Concepts used

basic proportionality theoremsimilar trianglesmedianprove

Approach

Step-by-step working

Concepts used

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