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Answer :

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

AB/PQ = AC/PR = AD/PM

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

AD = DE [By Construction.]

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 = AD/PM

CE/RN = AC/PR = 2AD/2PM

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

AB/PQ = AC/PR (Already given)

From equation (iii),

\(\angle\) A = \(\angle\) P

\(\triangle\) ABC ~ \(\triangle\) PQR [ SAS similarity criterion]

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