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In the diagram, △ABC and △ACD are drawn - NSC Mathematics - Question 9 - 2017 - Paper 2

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Question 9

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In the diagram, △ABC and △ACD are drawn. F and G are points on sides AB and AC respectively such that AF = 3x, FB = 2x, AG = 12y and GC = 8y. H, E and K are points o... show full transcript

Worked Solution & Example Answer:In the diagram, △ABC and △ACD are drawn - NSC Mathematics - Question 9 - 2017 - Paper 2

Step 1

FG || BC

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Answer

To prove that FG || BC, we can apply the concept of similar triangles.

From the geometric configuration, we know that:

  • AF : AG = 3x : 12y
  • FB : CG = 2x : 8y

This gives us the proportion: AFAG=FBCG\frac{AF}{AG} = \frac{FB}{CG} By the converse of the Basic Proportionality Theorem, since the line segments are proportional, it follows that FG || BC.

Step 2

AH / HK = AE / ED

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Answer

Applying the properties of parallel lines and similar triangles, we see:

  • Given that GH || CK, we can use the theorem of proportional segments.
  • This leads us to the proportion: AHHK=AEED\frac{AH}{HK} = \frac{AE}{ED} Thus, we can assert that the ratio of the segments AH to HK equals the ratio of AE to ED.

Step 3

If it is further given that AH = 15 and ED = 12, calculate the length of EK.

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Answer

We know that:

  • AH = 15 and ED = 12, with HK as the unknown variable.

Using the established ratio: 15HK=AE12\frac{15}{HK} = \frac{AE}{12} To solve for EK:

  1. Calculate AE using the ratios from similar triangles.
  2. From the geometric relationships, if AE is consistent with known values, it is derived by the relation EK = ED - HK.
  3. Solve for EK based on previously calculated lengths.

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