In the diagram, ΔABC is drawn. D is a point on AB and E is a point on AC such that DE || BC. BE and DC are drawn. Use the diagram to prove the theorem which states that a line drawn parallel to one side of a triangle divides the other two sides proportionally, in other words prove that AD/DB = AE/EC.

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In the diagram, ΔABC is drawn - NSC Mathematics - Question 10 - 2019 - Paper 2

Question 10

In the diagram, ΔABC is drawn. D is a point on AB and E is a point on AC such that DE || BC. BE and DC are drawn. Use the diagram to prove the theorem which states ... show full transcript

Worked Solution & Example Answer:In the diagram, ΔABC is drawn - NSC Mathematics - Question 10 - 2019 - Paper 2

Step 1

Draw h, from E to AD and h, from D to AE

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Answer

Construct a line h from point E to line AD, and another line h from point D to line AE, ensuring both lines are parallel to DE.

Step 2

Area equivalence

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Answer

By the properties of parallel lines, the area of triangle ADE can be expressed as:

$\text{area } ADE = \frac{1}{2} \times AD \times h_1$

And for triangle DBE:

$\text{area } DBE = \frac{1}{2} \times DB \times h_2$

Setting these areas equal gives:

$\text{area } ADE = \frac{AD}{DB} = \frac{AE}{EC}$

Step 3

Proportionality conclusion

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Answer

Since both areas are equal due to the same height and the proportionality of the bases being parallel, it follows that:

$\frac{AD}{DB} = \frac{AE}{EC}$

Thus proving that a line drawn parallel to one side of a triangle divides the other two sides proportionally.

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