CBSE 2025 · Region 2 · Set 3 · Q25 · 2 marks
In the given figure, $\displaystyle \frac{\mathrm{PS}}{\mathrm{SQ}}=\frac{\mathrm{PT}}{\mathrm{TR}}$ and $\displaystyle \angle \mathrm{PST}=\angle \mathrm{PRQ}$. Prove that $\displaystyle \Delta \mathrm{PQR}$ is an isosceles triangle.
In the given figure, $\displaystyle \Delta \mathrm{ABE} \cong \Delta \mathrm{ACD}$. Prove that $\displaystyle \Delta \mathrm{ADE} \sim \Delta \mathrm{ABC}$.
In the given figure, $\displaystyle \frac{\mathrm{PS}}{\mathrm{SQ}}=\frac{\mathrm{PT}}{\mathrm{TR}}$ and $\displaystyle \angle \mathrm{PST}=\angle \mathrm{PRQ}$. Prove that $\displaystyle \Delta \mathrm{PQR}$ is an isosceles triangle.
In the given figure, $\displaystyle \Delta \mathrm{ABE} \cong \Delta \mathrm{ACD}$. Prove that $\displaystyle \Delta \mathrm{ADE} \sim \Delta \mathrm{ABC}$.
Marking-scheme solution
Given \(\displaystyle \frac{\mathrm{PS}}{\mathrm{SQ}}=\frac{\mathrm{PT}}{\mathrm{TR}}\)
\[\begin{aligned}
& \Rightarrow \mathrm{ST} \| \mathrm{QR} \\
& \therefore \angle \mathrm{PST}=\angle \mathrm{PQR}
\end{aligned}
\]
and given, \(\displaystyle \angle \mathrm{PST}=\angle \mathrm{PRQ}\)
So, \(\displaystyle \angle \mathrm{PQR}=\angle \mathrm{PRQ}\)
\(\displaystyle \therefore \triangle \mathrm{PQR}\) is an isosceles triangle.
Given \(\displaystyle \Delta \mathrm{ABE} \cong \Delta \mathrm{ACD}\)
\[\therefore \mathrm{AE}=\mathrm{AD} \text { or } \mathrm{AD}=\mathrm{AE} \quad \text {---- ① }
\]
and \(\displaystyle \mathrm{AB}=\mathrm{AC} \quad\) ---- ②
Dividing ① by ②, we have
\[\frac{A D}{A B}=\frac{A E}{A C}
\]
and \(\displaystyle \angle \mathrm{DAE}=\angle \mathrm{BAC}\)
\[\therefore \triangle \mathrm{ADE} \sim \triangle \mathrm{ABC}
\]
TrianglesCriteria for Similarity of TrianglesAnalysevery_short_answerhard
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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.