E and F are points on the sides AB and AC respectively of a $\triangle ABC$ such that $\frac{AE}{EB} = \frac{AF}{FC} = \frac{1}{2}$. Which of the following relation is true ?
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
(b) BC = 2EF
Since $\dfrac{AE}{EB} = \dfrac{AF}{FC} = \dfrac{1}{2}$, by the converse of BPT, EF ∥ BC. Also, $\dfrac{AE}{AB} = \dfrac{1}{3}$. By AA similarity, △AEF ~ △ABC, so $\dfrac{EF}{BC} = \dfrac{AE}{AB} = \dfrac{1}{3}$, giving BC = 3EF.
Wait — re-checking: $AE:EB = 1:2$, so $AE:AB = 1:3$, hence $EF:BC = 1:3$, i.e., BC = 3EF.
Correct answer: (d) BC = 3EF
Source: Triangles, Section 6.3
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Explanation
- Since AE/EB = AF/FC = 1/2, by BPT converse EF ∥ BC, so △AEF ~ △ABC (AA criterion).
- The ratio AE : AB = AE : (AE + EB) = 1 : (1+2) = 1 : 3.
- Therefore EF/BC = 1/3, which means BC = 3EF → option (d).
- A common mistake is confusing AE/EB with AE/AB. Always convert the part-to-part ratio into a part-to-whole ratio before applying the similarity scale factor.