Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Thursday, November 12, 2015

### Geometry Problem 1164: Triangle, Angle Bisector, Cevian

Labels:
angle bisector,
cevian,
triangle

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Let AD=x so DE=2x, EC=6x and AB=3x

ReplyDeletecalculate BD and BC using cosine formula in triangles ABD and ABC .we get

BD^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/EC

so BE bisect angle DBC

Let AD = p so that DE = 2p, CE = 6p and AB = 3p

ReplyDeleteNow AB^2 = 9p^2 = AD.AC so AB is tangential to DBC at B

From similar Tr.s AD/AB = BD/BC = 1/3 = 2p/6p = DE/EC

Hence BE bisects < DBC

Sumith Peiris

Moratuwa

Sri Lanka

Tr BAE isoceles => ang ABE=ang AEB => ABD + DBE = EBC + ECB

ReplyDelete=> DBE = EBC ( ABD = ECB )

Excellent simple solution!

DeleteYes, but why ¿ABD=ECB?, because:

DeleteThe triangle ▲ABD is proportional to the triangle ▲BCA

BAC = BAD (They have in common the angle A)

Its sides are proportional:AC/AB=AB/AD=3

See relations between tang and secant

ReplyDeleteAng ABD = 1/2 arc BD, Ang ACB = 1/2 arc BD

or Tr ABD similar to Tr ABC

Use Stewarts Law with triangle ABC and the cevian BD:

ReplyDeleteAB = 3x, AD = x, DE = 2x, CE = 6x

(3x)^2*8x + BC^2*x = 9x(BD^2 + 2x^2)

That simplifies to BC^2 = 9BD^2 or BC = 3BD

Since BC:BD = 3:1 and CE:DE = 3:1 also BE is the angle bisector by the angle bisector theorem.