CBSE Class 10 Mathematics Standard 2025: question 33(b)

Hints

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Step-by-step solution

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1. Figure: Draw $△ABC$ with $BC$ as the base. Draw $AD\perp BC$ with foot $D$ on $BC$, so $\angle ADB=\angle ADC=90^{\circ }$.

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2. Given $A{D}^2=BD\times DC$. Dividing both sides by $AD\times DC$ gives
   $$\frac{BD}{AD}=\frac{AD}{DC}.$$

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3. In $△BDA$ and $△ADC$:

   • $\frac{BD}{AD}=\frac{AD}{DC}$ (shown above)
   • $\angle BDA=\angle ADC=90^{\circ }$ (included angles)

   So $△BDA\sim △ADC$ [SAS similarity criterion], with $B\leftrightarrow A$, $D\leftrightarrow D$, $A\leftrightarrow C$.

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4. Corresponding angles of similar triangles are equal:
   $$\angle DBA=\angle DAC\text{and}\angle BAD=\angle DCA.$$

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5. Adding these: $\angle DBA+\angle DCA=\angle DAC+\angle BAD=\angle BAC$.

   So $\angle B+\angle C=\angle BAC$.

   In $△ABC$, $\angle B+\angle C+\angle BAC=180^{\circ }$ [angle sum property].

   Substituting, $2\angle BAC=180^{\circ }$, so $\angle BAC=90^{\circ }$. Hence proved.

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Answer: $△BDA\sim △ADC$ (SAS), so $\angle B+\angle C=\angle BAC$, which gives $\angle BAC=90^{\circ }$. Proved.

Revise the chapter

• Chapter: Triangles (notes, formulas, where marks are lost and every past question)