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I have to proof in a triangle, that $\alpha_1>\alpha_2$ holds. The inner point P (from where I draw the smaller triangle) is set randomly.

enter image description here

Does anyone have a suggestion where I have to start?

Greetings

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You're probabaly missing one piece of vital information: The random point $P$ is supposed to be inside the original triangle? –  mrf Jun 5 '12 at 11:26
    
Yes, of course you're right –  ulead86 Jun 5 '12 at 11:30

3 Answers 3

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Given that the origin triangle is $\triangle ABC$, and the inner point is $P$, where $\alpha_1 = \angle BPC$, $\alpha_2 = \angle BAC$. Supposing that $D$ is the intersection of $AP$ and $BC$, we have $\angle BPC = \angle BPD + \angle DPC = \angle BAP + \angle ABP + \angle CAP + \angle ACP$ $= \angle BAC + \angle ABP + \angle ACP > \angle BAC$, Q.E.D.

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I tohught about this answer a few days but I can't find an explanation for $\angle BPD + \angle DPC = \angle BAP + \angle ABP + \angle CAP + \angle ACP$ perhaps you can give a hint? –  ulead86 Jun 8 '12 at 6:45
    
See here onlinemathlearning.com/exterior-angle-theorem.html, or have I made some mistakes? –  Frank Science Jun 8 '12 at 9:53

Each of $\,\alpha_1\,,\,\alpha_2\,$ equals $\,180 - (\beta_1+\beta_2)\,$ , with $\beta_i$ being the other two angles in the big (in the little red) triangle. As the other two angles of $\,\alpha_1\,$ are each less than the other two angles of $\,\alpha_2\,$ we get what we want.

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There is nothing more to do? I thought I have to argue, why the angles are smaller... –  ulead86 Jun 5 '12 at 11:27
    
@Daniels Of course you have! The other two angles of $\alpha_1$ are smaller than the corresponding ones of $\alpha_2$ since they're partial to them...or in other words: each of the other angles corres. to $\alpha_2$ equals the corres. angle of $\alpha_1$ plus a little more...you can mark this in your diagram, to make it clearer. –  DonAntonio Jun 5 '12 at 11:43

HINT: Draw a line that passes through the vertices of the two angles and the opposite side.

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