Proof for this theorem?

[miregal]

 

[lex]

For the first part, a<x+y - you could google "triangle inequality". You could attempt proving e.g. using the cosine rule.
For the second part: x+y<b+c, the problem is not fully specified. You need some extra information e.g. are [MATH]\angle [/MATH]PRC and [MATH]\angle [/MATH]PCR both acute angles or does [MATH]\bigtriangleup[/MATH]PRC have to be wholly within [MATH]\bigtriangleup[/MATH]ARC? etc...

 

[lex]

(If [MATH]\bigtriangleup[/MATH]PBC has to be wholly within [MATH]\bigtriangleup[/MATH]ABC (as it is drawn), then the second part can be proved using the triangle inequality).
I think I was misreading the B as a R.

 

[JeffM]

It is almost impossible to provide help on proofs because we do not what axioms and previous theorems are available to you.

 

[HallsofIvy]

First is should be obvious that angles PRC and RCP are smaller than angles ARC and ACR. Then apply trigonometry to right triangles ARQ and PCQ where Q is the foot of the perpendiculars from A and P to RC.

 

[miregal]

(If [MATH]\bigtriangleup[/MATH]PBC has to be wholly within [MATH]\bigtriangleup[/MATH]ABC (as it is drawn), then the second part can be proved using the triangle inequality).
I think I was misreading the B as a R.

I didn't know the theorem's name thanks for that. For the second part, because it is said that P is a point inside the triangle (sorry for not mentioning it), x+y will be less than b+c.

 

[lex]

Yes the extra information is needed to see what we can 'take for granted' from the picture that they have drawn.
You are right x+y<b+c, but the question is how might you prove it?

The proof must work for a set of triangles like this, containing obtuse angles:

(That's why it's important to know that P simply has to be within [MATH]\bigtriangleup[/MATH]ABC and that there is no special restriction on the angles).

 

[lex]

For the sake of completeness. Here is a proof:




and the argument clearly applies to a triangle with an obtuse angle too: