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

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

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## Wednesday, January 13, 2016

### Geometry Problem 1177: Triangle, Two Angle Bisectors, Metric Relations

Labels:
angle bisector,
metric relations,
triangle

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We can write 4 simple equations

ReplyDeletey+e = c.......(1)

x+d = a........(2)

c/b = x/d.......(3)

a/b = y/e.......(4)

Now we can eliminate a,b,c between these 4 equations.

From (3') & (4) dc/x = ae/y and so from (1) and (2)

d(y+e)/x = e(x+d)/y

Simplifying x^2e + xde + y^2d + yde

Sumith Peiris

Moratuwa

Sri Lanka

Manoloudis Apostolis 4 Lyceum of Korydallos-Greece

ReplyDeleteEven AB=c BC=a AC=b.Then x.e.(x+d)=y.d.(y+e) x.e,a=y.d.c

By theorem bisectors c/b=x/d and x=a.c/(b+c) d=a.b/(b+c) and a/b=y/e

y=a.c/(a+b) e=b.c/(a+b).

x.e.a=a.c/(b+c).b.c/(a+b).a=y.d.c .

apply two times bisector´s theorem

ReplyDeleteFirst for AD, and second for CE