The diagonal $BD$ of a parallelogram $ABCD$ intersects the line segment $AE$ at the point $F$, where $E$ is any point on the side $BC$. Prove that $DF \times EF = FB \times FA$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:21 · grounding rag
Model Answer
To Prove: $DF \times EF = FB \times FA$
Proof:
Consider $\triangle DFA$ and $\triangle BFE$.
In parallelogram $ABCD$, $AD \parallel BC$ (opposite sides).
Since $AD \parallel BC$ (i.e., $AD \parallel BE$), with $BD$ as transversal:
$$\angle FDA = \angle FBE \quad \text{(alternate interior angles)}$$
Also, with $AE$ as transversal:
$$\angle FAD = \angle FEB \quad \text{(alternate interior angles)}$$
Therefore, by AA similarity criterion:
$$\triangle DFA \sim \triangle BFE$$
Since corresponding sides of similar triangles are proportional:
$$\frac{DF}{BF} = \frac{FA}{FE}$$
Cross-multiplying:
$$DF \times FE = BF \times FA$$
$$\boxed{DF \times EF = FB \times FA} \qquad \textbf{(Proved)}$$
Source: Chapter 6 – Triangles, Section 6.4 Criteria for Similarity of Triangles (AA similarity criterion)
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Explanation
- Key step: Identify the two triangles — $\triangle DFA$ and $\triangle BFE$ — formed by the intersection of diagonal $BD$ and line segment $AE$ at point $F$.
- Why alternate angles? $AD \parallel BC$ in a parallelogram. Line $BD$ cuts these parallels → alternate angles $\angle FDA = \angle FBE$. Line $AE$ cuts these parallels → alternate angles $\angle FAD = \angle FEB$.
- AA criterion gives similarity; then the ratio of corresponding sides, when cross-multiplied, directly gives the required result.
- Examiners award marks for: correctly naming the triangles, stating both pairs of alternate angles with reason, writing the similarity statement, and the final cross-multiplication step.