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Mathematics · 2025 · 2 marks
CBSE 2025 · Region 5 · Set 3 · Q23
In ABC and PQR, AD and PS are altitudes such that ABD ∼ PQS and ACD ∼ PRS. Prove that ABC ∼ PQR.
In $\displaystyle \triangle \mathrm{ABC}$ and $\displaystyle \triangle \mathrm{PQR}, \mathrm{AD}$ and PS are altitudes such that $\displaystyle \triangle \mathrm{ABD} \sim \triangle \mathrm{PQS}$ and $\displaystyle \triangle \mathrm{ACD} \sim \triangle \mathrm{PRS}$. Prove that $\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$.
Marking-scheme solution
\[\Delta \mathrm{ABD} \sim \Delta \mathrm{PQS}
\]
\[\Longrightarrow \angle \mathrm{B}=\angle \mathrm{Q} \quad \text{--- (1)}
\]
\[\Delta \mathrm{ACD} \sim \Delta \mathrm{PRS}
\]
\[\Rightarrow \angle \mathrm{C}=\angle \mathrm{R} \quad \text{--- (2)}
\]
From ($\displaystyle 1$) & ($\displaystyle 2$), we get
\[\Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}
\]More from Triangles
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CBSE Class 10 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.