Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that AB/PQ = AC/PR = AD/PM


To Prove: ΔABC ~ ΔPQR


Construction: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.


Proof: In ΔABD and ΔCDE, we have

AD = DE  [By Construction]

BD = DC [∴ AP is the median]

and, ∠ADB = ∠CDE [Vertically opp. angles]

∴ ΔABD ≅ ΔCDE [By SAS criterion of congruence]

⇒ AB = CE [CPCT] ...(i)

Also, in ΔPQM and ΔMNR, we have

PM = MN [By Construction]

QM = MR [∴ PM is the median]

and, ∠PMQ = ∠NMR [Vertically opposite angles]

∴ ΔPQM = ΔMNR [By SAS criterion of congruence]

⇒ PQ = RN [CPCT] ...(ii)

Now, AB/PQ = AC/PR = AD/PM

⇒ CE/RN = AC/PR = AD/PM ...[From (i) and(ii)]

⇒ CE/RN = AC/PR = 2AD/2PM

⇒ CE/RN = AC/PR = AE/PN [∴ 2AD = AE and 2PM = PN]

∴ ΔACE ~ ΔPRN [By SSS similarity criterion]

Therefore, ∠2 = ∠4

Similarly, ∠1 = ∠3

∴ ∠1 + ∠2 = ∠3 + ∠4

⇒ ∠A = ∠P ...(iii)

Now, In ΔABC and ΔPQR, we have

AB/PQ = AC/PR (Given)

∠A = ∠P [From (iii)]

∴ ΔABC ~ ΔPQR [By SAS similarity criterion]

Triangle Proportional Sides and Medians

To prove that triangle ABC is similar to triangle PQR, we need to show that their corresponding sides are proportional. Given that sides AB and AC of triangle ABC are proportional to sides PQ and PR of triangle PQR, and median AD of triangle ABC is proportional to median PM of triangle PQR, we can use the triangle proportionality theorem to prove the similarity.

Triangle Proportionality Theorem

The triangle proportionality theorem states that if a line parallel to one side of a triangle intersects the other two sides, it divides those sides proportionally.

In our case, since sides AB and AC of triangle ABC are proportional to sides PQ and PR of triangle PQR, we can draw a line parallel to side BC of triangle ABC that intersects sides AB and AC at points E and F respectively. Similarly, we can draw a line parallel to side QR of triangle PQR that intersects sides PQ and PR at points N and O respectively.

Proving Proportional Sides

Using the triangle proportionality theorem, we can establish the following proportions:

1. AE/EB = PN/NQ (since AB/PQ = AC/PR)
2. AF/FC = PO/OR (since AC/PR = AB/PQ)
3. DE/EC = MN/NQ (since AD/PM = AB/PQ)
4. DF/FC = MO/OR (since AD/PM = AC/PR)

Proving Similarity

Now, let's consider triangles ADE and PNQ. We have established that AE/EB = PN/NQ and DE/EC = MN/NQ. By using the side-angle-side (SAS) similarity criterion, we can conclude that triangles ADE and PNQ are similar.

Similarly, considering triangles ADF and POR, we have DF/FC = MO/OR and AF/FC = PO/OR. By using the SAS similarity criterion, we can conclude that triangles ADF and POR are similar.

Conclusion

Since triangles ADE and PNQ are similar, and triangles ADF and POR are similar, we can conclude that triangles ABC and PQR are similar by the transitive property of similarity. Therefore, triangle ABC is similar to triangle PQR.