AC>AB in a triangle.D is a midpoint on AC such that AB=AD.Prove that CD<BC
B bills New member Joined Sep 5, 2007 Messages 1 Sep 6, 2007 #1 AC>AB in a triangle.D is a midpoint on AC such that AB=AD.Prove that CD<BC
D Deleted member 4993 Guest Sep 6, 2007 #2 Re: PROVE triangle inequality bills said: AC>AB in a triangle. D is a midpoint on AC such that AB=AD. Prove that CD<BC Click to expand... Please check your problem for accuracy - post the EXACT problem. Is D really the midpoint of AC - or is it just a point on AC? Please show your attempts to solve this problem - so that we know where to begin.
Re: PROVE triangle inequality bills said: AC>AB in a triangle. D is a midpoint on AC such that AB=AD. Prove that CD<BC Click to expand... Please check your problem for accuracy - post the EXACT problem. Is D really the midpoint of AC - or is it just a point on AC? Please show your attempts to solve this problem - so that we know where to begin.
D Denis Senior Member Joined Feb 17, 2004 Messages 1,707 Sep 6, 2007 #3 Re: PROVE triangle inequality: AC>AB, D midpt on AC; prov bills said: AC>AB in a triangle.D is a midpoint on AC such that AB=AD.Prove that CD<BC Click to expand... You then end up with AB=AD=DC; let each = x; then AB = x, AC = 2x Triangle law makes BC limits = (AC - AB + 1) to (AC + AB - 1), so: 2x - x + 1 to 2x + x - 1 x + 1 to 3x + 1 Since minimum BC = x + 1, and CD = x, then CD<BC : kapish?
Re: PROVE triangle inequality: AC>AB, D midpt on AC; prov bills said: AC>AB in a triangle.D is a midpoint on AC such that AB=AD.Prove that CD<BC Click to expand... You then end up with AB=AD=DC; let each = x; then AB = x, AC = 2x Triangle law makes BC limits = (AC - AB + 1) to (AC + AB - 1), so: 2x - x + 1 to 2x + x - 1 x + 1 to 3x + 1 Since minimum BC = x + 1, and CD = x, then CD<BC : kapish?
