Proof

[Thales12345]

Given:
triangle ABC with |AB|<|AC|
M is the midpoint of [BC]
P is the point on [AC] so that PM divides triangle ABC into 2 pieces with an equal perimeter
S is the intersection point of AB and PM

Proof that APS is isosceles

thanks for your help

 

[Dr.Peterson]

Given:
triangle ABC with |AB|<|AC|
M is the midpoint of [BC]
P is the point on [AC] so that PM divides triangle ABC into 2 pieces with an equal perimeter
S is the intersection point of AB and PM

Proof that APS is isosceles

thanks for your help

We like to help you solve problems yourself, and that requires us to have some idea of what help you need. Please show us any ideas you have, and any theorems you think might be useful.

I'll give you a start. The first thing I'd tell you to do is to make a picture (which also implies thinking about the meaning of the problem), so I'll make one for you:



I made this by deliberately placing P in the right place; can you see how I might do that? And it looks like the claim is correct, but of course we need a proof, not a "seems".

My next thought might be to try drawing another line that could produce a pair of similar or congruent triangles. I see a good place to do that. What ideas do you have?

 

[Thales12345]

I have already made a drawing and I was also thinking about congruent triangles. I made a parallelogram of it but I don’t think that’s it and now I’m stuck.

Maybe another option is working with a line segment bisector through M?

 

[Dr.Peterson]

Aim for similar triangles, not congruent. And don't add too much at once.

Try this parallel line:



Also, you might want to find some lengths. What are AP and PC in terms of a, b, and c? (That's what I used to construct my figure.)

 

[Thales12345]

Suppose that S is the intersection of b and that parallel line

angle MPS = angle QPA (opposite angles)
angle PMS = angle AQP (alternate interior angles)
so, if we can prove that PSM is isosceles, we can prove that AQP is isosceles.
but, I don’t find a way to show that |PS| = |MS| or that angle MPS = PMS ...

 

[Dr.Peterson]

Good so far.

Now use the fact that M is the midpoint to find CS and MS. (What special line is MS?)

Don't forget to do what I asked: What are AP and PC in terms of a, b, and c?

 

[Thales12345]

I don’t understand why you would use |CS|. I thought I was working with triangles PSM and AQP?
I can only see that MS is a parallel line to BA. Is MS a special line in other cases? I can’t find it.

|AP| + |PC| = b
|PC| - |AP| = c (I used this one to locate P)
Is this info necessary to come to the solution?

With Thales’ Theorem I’ve come to this, but I don’t know if it’s useful..
|CS|/|CA| = |CM|/|CB| = |MS|/|BA|

 

[Dr.Peterson]

I don’t understand why you would use |CS|. I thought I was working with triangles PSM and AQP?
I can only see that MS is a parallel line to BA. Is MS a special line in other cases? I can’t find it.

For my method, the important thing is to get |MS| and |PS|, which are part of the triangles of interest. What you'd really like to do is to show that PSM is isosceles, so that AQP also is.

Have you heard of a midline of a triangle? Or can you see that triangle SMC is similar to ABC?

|AP| + |PC| = b
|PC| - |AP| = c (I used this one to locate P)
Is this info necessary to come to the solution?

No one method is necessary; there are always alternatives. But have you tried solving these equations for |AP| and |PC|?

One of the first things I'd do with a problem like this is to try to label as many lengths as possible with expressions in a, b, c.

With Thales’ Theorem I’ve come to this, but I don’t know if it’s useful..
|CS|/|CA| = |CM|/|CB| = |MS|/|BA|

I haven't looked at your suggestion to see if it, too, can be useful.

 

[Thales12345]

Found it!

|PC| = |AP| + c

|MS| = c/2 #

|AS| = b/2

|AP| + |PC| = b
|AP| = b - |PC|
|AP| = b - (|AP| + c)
|AP| = b - |AP| - c
2|AP| = b - c
|AP| = b/2 - c/2

|AS| = |AP| + |PS|
b/2 = b/2 - c/2 +|PS|
|PS| = c/2 ##

# and ## : triangle PSM is isosceles (so APQ is isosceles too, I have already proved here above that triangle PSM is similar to triangle APQ)

thanks for your help

 

[Dr.Peterson]

Good. Your |AP| is how I constructed P to be at the right place, and the rest completes the proof you needed.