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Problem: Let A, B, C be three non-collinear points. Let D, E, F be points on the respective interiors of segments BC, AC and AB. Let θ, φ and ψ be the measures of the respective angles ∠BFC, ∠CDA and ∠AEB. Prove IAS(ABC) < θ +φ + ψ < 540 - IAS(ABC).(IAS means internal angle sum). Now im supposed to use the external angle inequality which is the measure of an exterior angle of a triangle is greater than that of either opposite interior angle. Not sure how to do it. Ive been struggling for hours with it. Oh i forgot to mention this is still in absolute geometry so we cant use that the the angles of a triangle add up to 180*.

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$\theta = \angle BFC \ge \angle BAC$, $\varphi = \angle CDA \ge \angle CBA$, etc? –  achille hui Feb 4 '13 at 5:06
    
umm ok but that doesnt help me too much. –  user60887 Feb 4 '13 at 5:08
    
$\psi = \angle AEB \ge \angle BCA$, add them up and isn't you get $\theta + \varphi + \psi \ge \operatorname{IAS}(ABC)$? –  achille hui Feb 4 '13 at 5:18
    
So I assume θ=∠BFC≥∠BAC, φ=∠CDA≥∠CBA, and ψ=∠AEB≥∠ACB OK so IAS(ABC)<= θ+φ+ψ. I get that part. –  user60887 Feb 4 '13 at 5:26
    
This is precisely the "external angle inequality" you mentioned. For the other part, look at the angles $\angle CFA, \angle ADB, \angle BEC$. You should draw everything on a piece of paper and view them as if you are doing Euclidean geometry, then the assignment of angles will become obvious. –  achille hui Feb 4 '13 at 5:33

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