Question

In the given figure, $D E ∥ A C$ and $D F ∥ A E$ . Prove that : $\frac{BF}{FE} = \frac{BE}{EC}$ .

Approach

In triangle ABC with DE ∥ AC and DF ∥ AE, use the Basic Proportionality Theorem twice. From DE ∥ AC in triangle BCA, $\frac{BF}{FE} = \frac{B D}{D A}$ . From ED ∥ CA in triangle BCA, $\frac{BE}{EC} = \frac{B D}{D A}$ . Hence $\frac{BF}{FE} = \frac{BE}{EC}$ .

Step-by-step working

1
In $△ BE A$ , since FD ∥ EA, by BPT: $\frac{BF}{FE} = \frac{B D}{D A}$ .
1 mark
2
In $△ BC A$ , since ED ∥ CA, by BPT: $\frac{BE}{EC} = \frac{B D}{D A}$ .
0.5 marks
3
From (1) and (2): $\frac{BF}{FE} = \frac{BE}{EC}$ .
0.5 marks

Concepts used

basic proportionality theoremparallel linesproveratio of sidesthales theorem

Approach

Step-by-step working

Concepts used

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