# Thread: Prove that triangle ABC is isosceles

1. ## 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.

2. Have you drawn a picture to help solve this problem? After doing this label the angles with the information given. Use the fact that a triangle has 180 degrees and the sum of angles around P is 360. Label your any unknowns (there should be 2) and solve them by setting up some simultaneous equations

3. i dont think i can use simultaneous equations to solve this.

4. 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: .$\displaystyle x + y \:=\:80^o$

5. I have drawn a sketch but don't know where to go from there. I think it's something to do with the cosine rule

6. This has been bothering me! If we don't crack it soon, I might seek "outside assistance"...

7. Is anyone listening?

With the given information, ANY triangle is possible.

(Well, with the exception of an equilateral.)

Here is a possible right triangle.

Code:

A
♥
*◊*
* ◊ *
*  ◊  *
*10 ◊ 40*
*    ◊    *
*     ♥     *
*    ◊ p      *
*20 ◊       ◊ 30*
*  ◊              *
* ◊               ◊ *
*◊ 70           10    *
B ♥  *  *  *  *  *  *  *  ♥ C

8. Soroban has already told you all you need to know- you can't prove it- it is NOT true!

9. I just wanted to see him draw another picture...

10. ## 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...

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