Sides AB and AC and median …

Given that triangle ABC and ΔPQR in which AD and PM are medians drawn on sides BC and QR respectively.

It is given that AB/PQ = AC/PR = AD/PM

We have to prove that ΔABC ~ ΔPQR

Construction: Produce AD to E such that AD = DE and PM to F such that PM = MF

From the figure,

In ΔABD and ΔCDE,

AD = DE               [by construction]

∠ADB = ∠CDE       [vertically opposite angles]

BD = DC              [Since AD is a median]

So, by SAS congruent condition

ΔABC ≅ ΔPQR

AB = CE               [by CPCT]

Similarly, we can prove

ΔPQM ≅ ΔRMF

PQ = RF               [by CPCT]

Now, given that

AB/PQ = AC/PR = AD/PM

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

CE/RF = AC/PR = AE/PF                     [Since AE = AD + DE and AD = DE, Same for PF]

By using SSS Congruent condition

ΔACE ≅ ΔPRF

=> ∠1 = ∠2   ......1

Similarly, ∠3 = ∠4    ......2

Adding equations 1 and 2, we get

∠1 + ∠3 = ∠2 + ∠4

=> ∠A = ∠P

Now, in ΔABC and ΔPQR

AB/PQ = AC/PR

and ∠A = ∠P

By SAS similar condtion,

ΔABC ~ ΔPQR