In the diagram above, ABC is a triangle. D and E are points on sides AB and AC respectively such that DE || BC. Use the diagram above to prove the theorem which states that a line drawn parallel to one side of a triangle divides the other two sides proportionally, i.e. prove that AD / DB = AE / EC.

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In the diagram above, ABC is a triangle - NSC Mathematics - Question 9 - 2021 - Paper 2

Question 9

In the diagram above, ABC is a triangle. D and E are points on sides AB and AC respectively such that DE || BC. Use the diagram above to prove the theorem which sta... show full transcript

Worked Solution & Example Answer:In the diagram above, ABC is a triangle - NSC Mathematics - Question 9 - 2021 - Paper 2

Step 1

Join BE and CD and draw h1 from E ⊥ AD and h2 from D ⊥ AE

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Answer

To begin the proof, we need to establish two perpendicular heights, h1 and h2, from points E and D respectively onto lines AD and AE. This forms right angles which will support our area comparisons.

Step 2

Establish the areas of triangles and apply the Area Proportionality Theorem

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Answer

The area of triangle ABC can be expressed using the height and base relationships. Since DE || BC, we have: $\frac{\text{area of } ADE}{\text{area of } ABC} = \frac{AD \cdot h1}{AB \cdot h}$ and $\frac{\text{area of } DEC}{\text{area of } ABC} = \frac{EC \cdot h2}{AC \cdot h}$ By the property of the ratios of areas, we conclude that: $\frac{AD}{DB} = \frac{AE}{EC}$

Step 3

Conclude the proof

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Answer

Since we have established the proportionality through the area ratios, we can conclude that: $\frac{AD}{DB} = \frac{AE}{EC}$ This completes the proof for the theorem that a line drawn parallel to one side of a triangle divides the other two sides proportionally.

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