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Mathematics · 2025 · 2 marks
CBSE 2025 · Region 2 · Set 1 · Q24
(a) In the given figure, D is a point on the side BC of Δ ABC such that ∠ ADC =∠ BAC. Show that CA^2= CD. CB.
In the given figure, D is a point on the side BC of $\displaystyle \Delta \mathrm{ABC}$ such that $\displaystyle \angle \mathrm{ADC}=\angle \mathrm{BAC}$. Show that $\displaystyle \mathrm{CA}^{2}=\mathrm{CD} . \mathrm{CB}$.
In the given figure, $\displaystyle \mathrm{OA} \cdot \mathrm{OB}=\mathrm{OC} \cdot \mathrm{OD}$. Show that $\displaystyle \angle \mathrm{A}=\angle \mathrm{C}$ and $\displaystyle \angle \mathrm{B}=\angle \mathrm{D}$.
Marking-scheme solution
In \(\displaystyle \Delta \mathrm{ACD}\) and \(\displaystyle \Delta \mathrm{BCA}\)
\[\begin{aligned}
& \angle \mathrm{ADC}=\angle \mathrm{BAC} \\
& \angle \mathrm{ACD}=\angle \mathrm{BCA}
\end{aligned}
\]
\(\displaystyle \therefore \triangle \mathrm{ACD} \sim \triangle \mathrm{BCA}\)
So, \(\displaystyle \frac{\mathrm{CA}}{\mathrm{CB}}=\frac{\mathrm{CD}}{\mathrm{CA}}\)
\(\displaystyle \Rightarrow \mathrm{CA}^{2}=\mathrm{CD} . \mathrm{CB}\)
Given \(\displaystyle \mathrm{OA} . \mathrm{OB}=\mathrm{OC} . \mathrm{OD}\)
\(\displaystyle \Rightarrow \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OD}}{\mathrm{OB}}\)
& \(\displaystyle \angle \mathrm{AOD}=\angle \mathrm{COB}\)
\(\displaystyle \therefore \triangle \mathrm{AOD} \sim \triangle \mathrm{COB}\)
So, \(\displaystyle \angle \mathrm{D}=\angle \mathrm{B}\) and \(\displaystyle \angle \mathrm{A}=\angle \mathrm{C}\)
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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.