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# Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that $$\triangle$$ ABC ~ $$\triangle$$ PQR.

Given: Two triangles $$\triangle$$ ABC and $$\triangle$$ PQR in which AD and PM are medians such that;

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

Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN. In $$\triangle$$ ABD and $$\triangle$$ CDE, we have
BD = DC [Since, AP is the median]
and, $$\angle$$ ADB = $$\angle$$ CDE
[Vertically opposite angles]

$$\triangle$$ ABD $$\cong$$ $$\triangle$$ CDE [SAS criterion of congruence]
AB = CE [By CPCT] …………………………..(i)

Also, in $$\triangle$$ PQM and $$\triangle$$ MNR,
PM = MN [By Construction.]
QM = MR [Since, PM is the median]
and, $$\angle$$ PMQ = $$\angle$$ NMR
[Vertically opposite angles]
$$\triangle$$ PQM = $$\triangle$$ MNR [SAS criterion of congruence]
PQ = RN [CPCT] ………………………………(ii)
Now, AB/PQ = AC/PR = AD/PM

From equation (i) and (ii),
CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]
$$\triangle$$ ACE ~ $$\triangle$$ PRN [SSS similarity criterion]

Therefore, $$\angle$$ 2 = $$\angle$$ 4
Similarly, $$\angle$$ 1 = $$\angle$$ 3
$$\angle$$ 1 + $$\angle$$ 2 = $$\angle$$ 3 + $$\angle$$ 4
$$\angle$$ A = $$\angle$$ P …………………………………………….(iii)

Now, In $$\triangle$$ ABC and $$\triangle$$ PQR, we have
$$\angle$$ A = $$\angle$$ P
$$\triangle$$ ABC ~ $$\triangle$$ PQR [ SAS similarity criterion]