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

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click the figure below to view more details of problem 1214.

## Wednesday, May 18, 2016

### Geometry Problem 1214: Triangle, Three Angle Bisectors, Proportions

Labels:
angle bisector,
proportions,
triangle

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c/a

ReplyDelete= AD/DC

= (AD/BD) / (DC/BD)

= ((c - e)/e) / ((a - f)/f)

ace - cef = acf - aef

1/f - 1/a = 1/e - 1/c

1/a + 1/e = 1/c + 1/f

c((a - f)/f) = a((c - e)/e)

Delete((a - f)/f)/a = ((c - e)/e)/c

1/f - 1/a = 1/e - 1/c

1/a + 1/e = 1/c + 1/f

Let AD = y; AC = x AND BD = z.

ReplyDeleteApplying angle Bisector to triangle ABC => x/y = a/c --------- (1)

Applying angle Bisector to triangle ABD => y/z = (c-e)/e ----------- (2)

Applying angle Bisector to triangle BDC => x/z = (a-f)/f -----------(3)

(3)/(2) => x/y = a/c = e.(a-f)/f.(c-e)

=> af(c-e) = ce(a-f)

=> a(c/e-1) = c(a/f-1)

=> 1/e-1/f = (a-c)/ac

=> 1/a+1?e = 1/c+1/f

1) c/AD=a/DC => 1.1)AD/DC=c/a

ReplyDelete2) (c-e)/AD=e/BD

3)(a-f)/DC=f/BD

from 2) & 3) 1.2) AD/DC=(c-e)f/(a-f)e

from 1.1 &1.2 it's proved