15 60

40 30 10

Analysis:

Initially, 60 can't be an angle. The reason for this is the following：

If 60 is in a corner, then there are two numbers on both sides of the corner, AB and CD.

Analyze the remaining numbers 10, 15, 20, 30, 40.

Take four, two are multiples of three, and there is only one left.

Therefore, it is inevitable to take multiples of 3 in ABCD.

And from AB=CD, we know that we should take two multiples of three, that is,15,30, which are not one side.

Then analyze the remaining 10, 20, 40. Only find it.

The combination of 40* 15=30*20, that is, the remaining 10 is filled on the opposite side of the corner.

But the product of each side is 40* 15*60=30*20*60=36000.

Then the product of the other two numbers on the opposite side of the angle should be 3600 except 10.

However, their maximum value can only be 30*40= 1200.

So to sum up, 60 can't be on the street corner, so 60 can only be on one side.

It is also considered that only 15, 30 and 60 are multiples of 3 in the number of 6.

And they can only be decomposed into a prime factor of 3, so15,30,60.

Only on three corners at the same time, or on three sides at the same time.

Otherwise, there will be a contradiction. The product of three numbers on one side is a multiple of 9 or even 27, while at least one on the other side can only decompose a prime factor of 3. It is equal to the product of three numbers on both sides.

According to the above conclusion, we know that 15, 30 and 60 are on three sides of the triangle.

Now draw a triangle ABC, D on BC, E on AC and F on AB.

Let d be 15, E 30, f and f 60.

We know from AFB=BDC that A:C=D:F= 1:4.

10,20,40 can only be found in 10:40= 1:4, so a is 10 and c is 40. So B=20.

The substitution verification is correct, and this filling method can be used.

In fact, there is essentially only one filling method.

Other filling methods are obtained through this cyclic exchange or symmetry.