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Question

If $$AD$$ and $$PM$$ are medians of triangles $$ABC$$ and $$PQR,$$ respectively where $$\triangle ABC \sim \triangle PQR$$, prove that $$\dfrac{AB}{PQ}=\dfrac{AD}{PM}$$.
465445_44875788618f4e3b8ecf126f4f57188c.png


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 Proved

494420_465445_ans_369dc4acd15943d499b9efc62bb4126b.png

Mathematics
RS Agarwal
Standard X

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