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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 4 · Set 2 · Q22
In the given figure, DE ∥ AC and DF ∥ AE. Prove that: BF/FE=BE/EC.
In the given figure, $\displaystyle \mathrm{DE} \| \mathrm{AC}$ and $\displaystyle \mathrm{DF} \| \mathrm{AE}$. Prove that : $\displaystyle \frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}$.
Marking-scheme solution
In \(\displaystyle \triangle \mathrm{BEA}, \mathrm{FD} \| \mathrm{EA}\)
\[\therefore \frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BD}}{\mathrm{DA}}
\]
In \(\displaystyle \triangle \mathrm{BCA}, \mathrm{ED} \| \mathrm{CA}\)
\[\therefore \frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BD}}{\mathrm{DA}}
\]
Using (i) and (ii)
\[\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{E C}
\]
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CBSE Class 10 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.