CBSE 2025 · Region 6 · Set 1 · Q24 · 2 marks
P is a point on the side BC of $\displaystyle \triangle \mathrm{ABC}$ such that $\displaystyle \angle \mathrm{APC}=\angle \mathrm{BAC}$. Prove that $\displaystyle \mathrm{AC}^{2}=\mathrm{BC} \cdot \mathrm{CP}$.
Marking-scheme solution
\[\angle \mathrm{APC}=\angle \mathrm{BAC}
\]
\[\angle \mathrm{ACP}=\angle \mathrm{ACB}
\]
\[\therefore \Delta \mathrm{APC} \sim \Delta \mathrm{BAC}
\]
\[\mathrm{AC}^{2}=\mathrm{BC} \cdot \mathrm{CP}
\]TrianglesApplications of SimilarityAnalysevery_short_answermedium
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.