Hello, ahm0605!

$\displaystyle \text}Triangle }ABC\text{ has the following properties:}$

$\displaystyle \text{There is an interior point }P\text{ such that:}$

. . $\displaystyle \angle PAB = 10^o,\;\angle PBA = 20^o, \;\angle PCA = 30^o,\;\angle PAC = 40^o.$

$\displaystyle \text{Prove that triangle }ABC\text{ is isosceles.}$

Very difficult problem, don't know where to start.

Of course you do . . . Make a sketch!

I don't think the proof is possible.

Code:

A
o
*
***
*
* * *
*
* * *
10 * 40
* * *
*
* * *
*
* * *
*
* o *
150 * 110
* * * *
20 * 100 * 30
* * * *
* x y *
B o * * * * o C

And all we have is: .$\displaystyle x + y \:=\:80^o$