CBSE Class 10 Mathematics Basic 2024: question 32(B)

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1. Given $\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$, AD and PM medians. Construction: produce AD to E with $DE=AD$, join CE; produce PM to N with $MN=PM$, join RN. Mark $\angle BAD=\angle 3$, $\angle DAC=\angle 1$, $\angle QPM=\angle 4$, $\angle MPR=\angle 2$.

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2. In $△ADB$ and $△EDC$: $BD=DC$ [D midpoint], $AD=DE$ [construction], $\angle ADB=\angle EDC$ [vertically opposite]. So $△ADB\cong △EDC$ (SAS) $\Rightarrow AB=CE$ and $\angle 3=\angle CED$. Similarly $△PMQ\cong △NMR\Rightarrow PQ=RN$ and $\angle 4=\angle RNM$.

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3. Now $\frac{CE}{RN}=\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}=\frac{2AD}{2PM}=\frac{AE}{PN}$. Hence $△AEC\sim △PNR$ [SSS similarity].

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4. So $\angle 1=\angle 2$ (i.e. $\angle CAE=\angle RPN$) and $\angle AEC=\angle PNR$, giving $\angle 3=\angle CED=\angle RNM=\angle 4$. Adding: $\angle 1+\angle 3=\angle 2+\angle 4\Rightarrow \angle BAC=\angle QPR$.

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5. In $△ABC$ and $△PQR$: $\frac{AB}{PQ}=\frac{AC}{PR}$ [given] and $\angle BAC=\angle QPR$. Therefore $△ABC\sim △PQR$ [SAS similarity].

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Answer: $△ABC\sim △PQR$ (SAS similarity, after showing $\angle BAC=\angle QPR$)

Revise the chapter

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