Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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*.

share|cite|improve this question
$\theta = \angle BFC \ge \angle BAC$, $\varphi = \angle CDA \ge \angle CBA$, 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.