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Mathematics · 2024 · 5 marks
CBSE 2024 · Region 4 · Set 1 · Q33
(a) E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ∼ Δ CFB.
(a)
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F . Show that $\displaystyle \Delta \mathrm{ABE} \sim \Delta \mathrm{CFB}$.
OR
(b)
Sides AB, BC and the median AD of $\displaystyle \Delta \mathrm{ABC}$ are respectively proportional to sides $\displaystyle \mathrm{PQ}, \mathrm{QR}$ and the median PM of another $\displaystyle \Delta \mathrm{PQR}$. Prove that $\displaystyle \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$.
Marking-scheme solution
In \(\displaystyle \triangle \mathrm{ABE}\) and \(\displaystyle \triangle \mathrm{CFB}\)
\[\begin{aligned}
& \angle \mathrm{EAB}=\angle \mathrm{BCF} \\
& \angle \mathrm{AEB}=\angle \mathrm{CBF} \\
& \Rightarrow \triangle \mathrm{ABE} \sim \triangle \mathrm{CFB}
\end{aligned}
\]
\[\begin{aligned}
& \because \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M} \\
& \therefore \frac{A B}{P Q}=\frac{2 B D}{2 Q M}=\frac{A D}{P M}
\end{aligned}
\]
\[\begin{aligned}
& \Rightarrow \frac{A B}{P Q}=\frac{B D}{Q M}=\frac{A D}{P M} \\
& \Rightarrow \triangle \mathrm{ABD} \sim \triangle \mathrm{PQM} \\
& \Rightarrow \angle \mathrm{~B}=\angle \mathrm{Q}
\end{aligned}
\]
In \(\displaystyle \triangle \mathrm{ABC}\) and \(\displaystyle \triangle \mathrm{PQR}\)
\[\begin{aligned}
& \frac{A B}{P Q}=\frac{B C}{Q R} \\
& \angle \mathrm{~B}=\angle \mathrm{Q} \\
& \therefore \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.