# Prove that triangle ABC is isosceles

• Mar 13th 2011, 12:58 PM
ahm0605
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.
• Mar 13th 2011, 01:07 PM
pickslides
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
• Mar 13th 2011, 01:34 PM
ahm0605
i dont think i can use simultaneous equations to solve this.
• Mar 14th 2011, 09:13 AM
Soroban
Hello, ahm0605!

Quote:

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

$\displaystyle \text{There is an interior point }P\text{ such that:}$
. . $\displaystyle \angle PAB = 10^o,\;\angle PBA = 20^o, \;\angle PCA = 30^o,\;\angle PAC = 40^o.$

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

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

• Mar 14th 2011, 07:37 PM
ahm0605
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
• Mar 14th 2011, 07:45 PM
TheChaz
This has been bothering me! If we don't crack it soon, I might seek "outside assistance"...
• Mar 15th 2011, 04:20 AM
Soroban

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
• Mar 15th 2011, 04:27 AM
HallsofIvy
Soroban has already told you all you need to know- you can't prove it- it is NOT true!
• Mar 15th 2011, 05:05 AM
TheChaz
I just wanted to see him draw another picture...
(Clapping)
(Heart)
• Jan 25th 2013, 10:48 AM
earthboy
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...
consider $\displaystyle \angle PCB = x$ and we know $\displaystyle \angle PBC + \angle PCB =80$, hence$\displaystyle \angle PBC= 80-x$
$\displaystyle \frac{\sin \angle PBA}{\sin \angle PAB}*\frac{\sin \angle PCB}{\sin \angle PBC}*\frac{\sin \angle PAC}{\sin \angle PCA} =1$
=> $\displaystyle \frac{\sin 20 \sin x \sin 40}{\sin 10 \sin (80-x) \sin 30}=1$
=>$\displaystyle \frac{4\sin x \sin 40 \cos 10}{\sin (80-x)}=1$
=>$\displaystyle \frac{2\sin x(\sin 30+\sin 50)}{\sin (80-x)}=1$ [#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