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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 4 · Set 1 · Q33
D is a point on the side BC of a triangle ABC such that $\displaystyle \angle \mathrm{ADC}=\angle \mathrm{BAC}$, prove that $\displaystyle \mathrm{CA}^{2}=\mathrm{CB} . \mathrm{CD}$If AD and PM are medians of triangles ABC and PQR , respectively where $\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$, prove that $\displaystyle \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}$.
D is a point on the side BC of a triangle ABC such that $\displaystyle \angle \mathrm{ADC}=\angle \mathrm{BAC}$, prove that $\displaystyle \mathrm{CA}^{2}=\mathrm{CB} . \mathrm{CD}$
If AD and PM are medians of triangles ABC and PQR , respectively where $\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$, prove that $\displaystyle \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}$.
Marking-scheme solution
In \(\displaystyle \triangle \mathrm{ABC}\), D is a point on side BC such that \(\displaystyle \angle \mathrm{ADC}=\angle \mathrm{BAC}\) In \(\displaystyle \Delta \mathrm{CBA}\) and \(\displaystyle \Delta \mathrm{CDA}\)
\[\begin{aligned}
& \angle \mathrm{C}=\angle \mathrm{C} \text { (common) } \\
& \angle \mathrm{BAC}=\angle \mathrm{ADC} \text { (given) } \\
& \therefore \triangle \mathrm{CBA} \sim \triangle \mathrm{CAD} \text { (By AA similarity) } \\
& \therefore \text { their corresponding sides are proportional } \\
& =\frac{\mathrm{CA}}{\mathrm{CD}} \Rightarrow \mathrm{CA}^{2}=\mathrm{CB} \cdot \mathrm{CD}
\end{aligned}
\]
\(\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}\)
AD and AM are medians of \(\displaystyle \Delta \mathrm{ABC}\) and \(\displaystyle \Delta \mathrm{PQR}\) respectively.
\(\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}\)
\[\therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}
\]
\(\displaystyle \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{2 \mathrm{BD}}{2 \mathrm{QM}}\)
\(\displaystyle \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}\)
Also \(\displaystyle \angle \mathrm{B}=\angle \mathrm{Q} \quad(\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR})\)
\(\displaystyle \Rightarrow \Delta \mathrm{ABD} \sim \Delta \mathrm{PQM}\) (SAS similarly)
\(\displaystyle \Rightarrow \frac{\boldsymbol{A} \boldsymbol{B}}{\boldsymbol{P} \boldsymbol{Q}}=\frac{\boldsymbol{A} \boldsymbol{D}}{\boldsymbol{P} \boldsymbol{M}}\)
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CBSE Class 10 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.