Prove that triangle ABC is isosceles

Triangle ABC has the following properties: there is an interior point P such that angle PAB = 10 degrees, angle PBA = 20 degrees, angle PCA = 30 degrees and angle PAC = 40 degrees. Prove that triangle ABC is isosceles.

Very difficult problem don't know where to start.

Any help appreciated.

Re: Prove that triangle ABC is isosceles

Sorry to open the old thread, but a user has just posted this problem recently, and I think i should give a link to the solution, as the problem is both famous, interesting and not wrong...

mathhelpforum.com/new-users/212009-mongolias-province-math-competition-9th-grade.html#post765883

the link does not seem to work:

consider $\displaystyle \angle PCB = x$ and we know $\displaystyle \angle PBC + \angle PCB =80$, hence$\displaystyle \angle PBC= 80-x$

now we have to use the trigonometric version of ceva's theorem:

$\displaystyle \frac{\sin \angle PBA}{\sin \angle PAB}*\frac{\sin \angle PCB}{\sin \angle PBC}*\frac{\sin \angle PAC}{\sin \angle PCA} =1$

=> $\displaystyle \frac{\sin 20 \sin x \sin 40}{\sin 10 \sin (80-x) \sin 30}=1$

=>$\displaystyle \frac{4\sin x \sin 40 \cos 10}{\sin (80-x)}=1$

=>$\displaystyle \frac{2\sin x(\sin 30+\sin 50)}{\sin (80-x)}=1$ [#using product-sum formula]

=>$\displaystyle \frac{\sin x(1+2\cos 40)}{\sin (80-x)}=1$

=>$\displaystyle 2\sin x \cos 40 = \sin (80-x)-\sin x = 2\sin(40-x)\cos 40$

so as $\displaystyle 2\sin x \cos 40 = 2\sin(40-x)\cos 40$, we can write $\displaystyle x=40-x$ or $\displaystyle x=20$

hence $\displaystyle \angle ACB=\angle ABC = 50$,therefore triangle ABC is isosceles