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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below for more details.
Let AD=x so DE=2x, EC=6x and AB=3xcalculate BD and BC using cosine formula in triangles ABD and ABC .we getBD^2=10x^2-6x^2.cos(A)BC^2=90x^2-54x^2.cos(A)(BD/BC)^2= 1/9 => BD/BC=1/3= ED/ECso BE bisect angle DBC
Let AD = p so that DE = 2p, CE = 6p and AB = 3pNow AB^2 = 9p^2 = AD.AC so AB is tangential to DBC at BFrom similar Tr.s AD/AB = BD/BC = 1/3 = 2p/6p = DE/EC Hence BE bisects < DBCSumith PeirisMoratuwaSri Lanka
Tr BAE isoceles => ang ABE=ang AEB => ABD + DBE = EBC + ECB => DBE = EBC ( ABD = ECB )
Excellent simple solution!
Yes, but why ¿ABD=ECB?, because: The triangle ▲ABD is proportional to the triangle ▲BCABAC = BAD (They have in common the angle A)Its sides are proportional:AC/AB=AB/AD=3
See relations between tang and secantAng ABD = 1/2 arc BD, Ang ACB = 1/2 arc BDor Tr ABD similar to Tr ABC