Given a triangle ABC with ∠ B = 60°. The bisectors of angle A and C intersect BC and AB at D and E respectively. Prove that CD + AE = AC.
Please help me with this.... I tried hours but still can't prove it.
Using angle bisector theorem, i get (EB/AC)= (AE)/(AC) and (BD)/(DC) = (AB)/AC). So by combining the results, i get
(AE + DC) / (AC) = [ (EB)/BC)] + [(BD)/(AB)]
(AE + DC) = (AC) [(EB)/BC)] + [(BD)/(AB)]
Am i in the correct path? How should i proceed?
Michael, is this homework...or what?
Did you "kinda see what's going on" by drawing a representative triangle?
Anyway, I'll give you this, then you're on your own:
M is the point where the bisectors cross...OK?Code:B(60) c-e a-d E D M e d C(2u) b A(120-2u)
Let a = BC, b = AC, c = AB, d = CD and e = AE; then BD = a-d and BE = c-e.
(Easier to work, as example, with "b" instead of "AC"...get my drift?)
So the ratios are: a / (c+e) = b / e and c / (a-d) = b / d
OK: let 2u = angleACB ; then angleACE = u.
Also, angle BAC becomes 120-2u ; then angleCAD = 60-u
Now using triangleAMC, angleAMC = 180-u-(60-u) = 120 : see that?
This forces angleAME = 60 (also angleCMD)
Carry on...
Its simple...
Let the point of intersection of AD and CE be O
draw the angle bisector of O meeting AC at M.
Now, LCOD = 60 so, LCOM = 60 = LMOA = LAOE
Thus, Triangles COD and COM && Triangles AOM and AOE are congruent..
Thus, AE = AM and CD = CM or, CD + AE = AM + CM = AC. Done...!!
Hmmmm...well, true Burnit...but we usually do not give full solution here.
I think Michael would have been able to wrap this up from my hints:
he seems to know what he's doing...
Showing him that the "trick" is to arrive at central angle = 120 should then
make it evident to him that next step is to create these congruent triangles.
But then who knows...he hasn't responded !