Question

# If $$AD$$ and $$PM$$ are medians of triangles $$ABC$$ and $$PQR,$$ respectively where $$\triangle ABC \sim \triangle PQR$$, prove that $$\dfrac{AB}{PQ}=\dfrac{AD}{PM}$$.

Solution

## Since, $$\triangle ABC$$ $$\sim$$ $$\triangle PQR$$$$\therefore$$ $$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AC}{PR}$$$$\angle A=\angle P\,,\angle B=\angle Q\,,\angle C=\angle R$$But, $$BC=2BD$$ and $$QR=2QM$$Hence, $$\dfrac{AB}{PQ}=\dfrac{BD}{QM}=\dfrac{AC}{PR}$$In $$\triangle ABD$$ and $$\triangle PQM,$$$$\dfrac{AB}{PQ}=\dfrac{BD}{QM}$$ and $$\angle B=\angle Q$$By SAS similarity, $$\triangle ABD$$ $$\sim$$ $$\triangle PQM,$$$$\therefore$$ $$\dfrac{AB}{PQ}=\dfrac{AD}{PM}$$Hence ProvedMathematicsRS AgarwalStandard X

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