Hi, I need any kind of help I can get right now.. My friend and I are really stuck on this question...
In any triangle ABC, E and D are interior points of AC and BC respectively. AF bisects angle CAD and BF bisects angle CBE. Prove that m(angle AEB) + m(angle ADB) = 2m(angle AFB).
m is just the measure of the angle..
2007-02-01 18:27:01 · 4 answers · asked by yogastar02 2 in Science & Mathematics ➔ Mathematics
That is exactly how the question is worded... we're not given a picture, we have to draw it ourselves...
2007-02-01 18:45:24 · update #1
This is straightfoward. aEAF = aFAD and aDBF = aFBC by statement of the problem, AF and BF being bisectors. We know then that:
180 = 2 aEAF + aDAB + aEBA+ aAEB
180 = 2 aEBF + aEBA + aDAB+ aADB, so that we know that
aAEB + aADB = 360 - 2 aEAF - 2 aDAB - 2 aEBF - 2 aEBA
but we also know that in triange ABF,
180 = aEAF + aDAB + aEBF + aEBA + aAFB, or doubling it and rearranging,
360 - 2 aEAF - 2 DAB - 2 aEBF - 2 EBA = 2 aAFB
Hence, aAEB + aADB = 2 aAFB
To draw the picture, first draw triangle ABC, with C at top, then mark point F somewhere inside of it, and draw triangle ABF. Then draw AD and BE such that lines AF and BF are bisectors of aCAD and aCBE
2007-02-01 19:08:43 · answer #1 · answered by Scythian1950 7 · 0⤊ 0⤋
I'm doing geometry and trigonometry as well but you will have to ask someone who can actually see the question. It's too difficult to understand and solve without looking at it. Sorry.
2007-02-02 02:43:39 · answer #2 · answered by Anonymous · 0⤊ 0⤋
If you could e-mail a picture of it to me I could solve it for you
2007-02-02 02:51:02 · answer #3 · answered by snilubez 2 · 0⤊ 0⤋
it is to hard to understand from what you are telling me.
Ask someone for help that can actuall see the question
Sorry
2007-02-02 02:38:05 · answer #4 · answered by donkeywollenjumper 1 · 0⤊ 0⤋