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Two bisectors BD and CE of triangle ABC intersect at point O, angle a = 60 degrees, proving: CD + be = BC There is an answer to this question on the Internet, but I can't understand it because of my low IQ. Dashen explains why ∠ EOB = 1 / 2 (180 ° - 60 °) = 60 ° This is a certain method: take a point F on the edge of BC, make BF = be, link of ∵ BD is the angular bisector, BF = be, Bo is the common edge, ∴△BEO≌△BFO →∠EOB＝∠FOB＝∠COD ∵∠A＝60° The angle bisector is BD and CE ∵∠EOB＝1/2（180°－60°）＝60° Then ∠ COF = 180 ° - ∠ FOB - ∠ cod = 60 ° And ∵ CE is the angular bisector and CO is the common edge, ∠ COF = ∠ cod = 60 ° (proved) ∴△COF≌△COD →CF＝CD ∴BC＝BF＋CF＝BE＋CD Or there are other ways,
If the bisectors be and CF of △ ABC intersect at one point I, then ∠ BIC = 90 ° + A / 2. If you recite this conclusion for me, you will not prove it. Please recite it
So in this question, ∠ BOC = 90 °+ ∠ A / 2 = 120 °. ∠ EOC = 180 °- ∠ BOC = 60 °
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