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

Given, $$\triangle$$ ABC and $$\triangle$$ PQR, AB, BC and median AD of $$\triangle$$ ABC are proportional to sides PQ, QR and median PM of $$\triangle$$ PQR

i.e. AB/PQ = BC/QR = AD/PM

We have to prove: $$\triangle$$ ABC ~ $$\triangle$$ PQR

As we know here,
$${{AB} \over {PQ}} = {{{{1} \over {2}}BC} \over {{{1} \over {2}}QR}} = {{AD} \over {PM}}$$
(D is the midpoint of BC. M is the midpoint of QR)
$$\triangle$$ ABD ~ $$\triangle$$ PQM [SSS similarity criterion]

$$\angle$$ ABD = $$\angle$$ PQM
[Corresponding angles of two similar triangles are equal]
$$\angle$$ ABC = $$\angle$$ PQR

In $$\triangle$$ ABC and $$\triangle$$ PQR
AB/PQ = BC/QR ………………………….(i)
$$\angle$$ ABC = $$\angle$$ PQR ……………………………(ii)

From equation (i) and (ii), we get,
$$\triangle$$ ABC ~ $$\triangle$$ PQR [SAS similarity criterion]