**Proof:**

Given, ΔABC ~ ΔPQR

We need to prove AB/PQ = AD/PM

**Step 1:**

Let's first understand the concept of medians.

A median of a triangle is a line segment that connects a vertex of the triangle to the midpoint of the opposite side.

In triangle ABC, AD is a median. So, it passes through the midpoint of BC.

Similarly, in triangle PQR, PM is a median. So, it passes through the midpoint of QR.

**Step 2:**

Since ΔABC ~ ΔPQR, we know that the corresponding sides of the two triangles are proportional.

So, we can write:

AB/PQ = BC/QR

**Step 3:**

Now, let's look at the triangles formed by the medians.

In triangle ABC, let M be the midpoint of AD.

Similarly, in triangle PQR, let N be the midpoint of PM.

**Step 4:**

We know that the medians of a triangle divide it into six smaller triangles of equal area.

So, we can write:

Area(ΔABM) = Area(ΔACM) = 1/6 Area(ΔABC)

Area(ΔPNM) = Area(ΔPQM) = 1/6 Area(ΔPQR)

**Step 5:**

Let's now look at the ratio of the areas of the two triangles ΔABM and ΔPNM.

We know that the corresponding sides of the two triangles are proportional. So, we can write:

AB/PQ = BC/QR = AC/PR

Therefore, we have:

AB/AC = PQ/PR

**Step 6:**

We can now use the formula for the area of a triangle, which is:

Area = 1/2 base × height

In triangle ΔABM, the height is AD and the base is AB.

So, we have:

Area(ΔABM) = 1/2 × AB × AD

Similarly, in triangle ΔPNM, the height is PM and the base is PQ.

So, we have:

Area(ΔPNM) = 1/2 × PQ × PM

**Step 7:**

We can now substitute the values of the areas in terms of the bases and heights and simplify the expression.

We get:

AB/PQ = (2 × Area(ΔABM))/AD × (1/2)/(2 × Area(ΔPNM))/PM × (1/2)

AB/PQ = (Area(ΔABM))/AD × (1/Area(ΔPNM))/PM

AB/PQ = AD/PM

Hence, proved.

Therefore, AB/PQ = AD/PM.