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) 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

Given: In triangle ABC, a line DE is drawn parallel to side BC, intersecting AB and AC at D and E respectively. To prove: $\frac{A D}{D B} = \frac{A E}{EC}$ . Construction: Draw perpendiculars DM to AC and EN to AB. Proof: Area ratios give $\frac{a r ( △ D BE )}{a r ( △ A D E )} = \frac{D B}{A D}$ and $\frac{a r ( △ D CE )}{a r ( △ A D E )} = \frac{EC}{A E}$ . Since DE || BC, triangles DBE and DCE have equal areas (same base, same height). Hence $\frac{A D}{D B} = \frac{A E}{EC}$ .

Step-by-step working

1
Given: In $△ A BC$ , $D E ∥ BC$ . To prove: $\frac{A D}{D B} = \frac{A E}{EC}$
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2
Construction: Draw $D M ⊥ A C$ and $EN ⊥ A B$ . Join BE and CD
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3
$\frac{a r ( △ D BE )}{a r ( △ A D E )} = \frac{\frac{1}{2} \cdot D B \cdot EN}{\frac{1}{2} \cdot A D \cdot EN} = \frac{D B}{A D}$
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4
$\frac{a r ( △ D CE )}{a r ( △ A D E )} = \frac{\frac{1}{2} \cdot EC \cdot D M}{\frac{1}{2} \cdot A E \cdot D M} = \frac{EC}{A E}$
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5
Since DE || BC, triangles DBE and DCE lie on the same base DE and between the same parallels, so $a r (△ D BE) = a r (△ D CE)$
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6
Therefore, $\frac{a r ( △ D BE )}{a r ( △ A D E )} = \frac{a r ( △ D CE )}{a r ( △ A D E )} \Rightarrow \frac{D B}{A D} = \frac{EC}{A E}$
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7
Hence, $\frac{A D}{D B} = \frac{A E}{EC}$
1 mark

Concepts used

basic proportionality theoremsimilar trianglesprovemedianratio

Approach

Step-by-step working

Concepts used

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