Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.

Asked by Abhisek | 1 year ago |  96

##### Solution :-

Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR

i.e. $$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PM}$$

We have to prove: ΔABC ~ ΔPQR

As we know here,

$$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PM}$$

$$\dfrac{AB}{PQ}=\dfrac{\dfrac{1}{2}BC}{\dfrac{1}{2}QR}=\dfrac{AD}{PM}$$............(i)

$$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PM}$$ (D is the midpoint of BC. M is the midpoint of QR)

ΔABD ~ ΔPQM [SSS similarity criterion]

∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]

⇒ ∠ABC = ∠PQR

In ΔABC and ΔPQR

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

∠ABC = ∠PQR ……………… ……(ii)

From equation (i) and (ii), we get,

ΔABC ~ ΔPQR [SAS similarity criterion]

Answered by Pragya Singh | 1 year ago

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