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

Asked by Abhisek | 1 year ago |  84

##### Solution :-

Given, ΔABC ~ ΔPQR

We know that the corresponding sides of similar triangles are in proportion.

$$\dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{BC}{QR}$$……(i)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ……..(ii)

Since AD and PM are medians, they will divide their opposite sides.

∴ BD = $$\dfrac{BC}{2}$$ and QM =$$\dfrac{QR}{2}$$  ……….(iii)

From equations (i) and (iii), we get

$$\dfrac{AB}{PQ}=\dfrac{BD}{QM}$$ ………….(iv)

In ΔABD and ΔPQM,

From equation (ii), we have

∠B = ∠Q

From equation (iv), we have,

$$\dfrac{AB}{PQ}=\dfrac{BD}{QM}$$

ΔABD ~ ΔPQM (SAS similarity criterion)

⇒ $$\dfrac{AB}{PQ}=\dfrac{BD}{QM}$$$$\dfrac{AD}{PM}$$

Answered by Pragya Singh | 1 year ago

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