Posted by Mona on Friday, February 1, 2013 at 12:12pm.
The perimeter of triangle ABC is 120. LIne segment BD bisects <B. Line segment AD= 15, and line segment DC= 21. FInd AB and BC.
Thanks.

Geometry8th gr  Reiny, Friday, February 1, 2013 at 2:24pm
This is a very tough question for grade 8, I am totally impressed.
make your sketch, let
AB = a
BC = b
BD = x
Angle B = 2Ø , so that angle ABD = Ø and angle BCD = Ø
I am going to use the fact that if we have triangle with sides a and b with Ø the angle between them
the area = (1/2)ab sinØ
area of triangle ABD= (1/2)ax sinØ
area of triangle BCD=(1/2)bx sinØ
But if we consider the bases along AC , the both have the same measured from B
so they are in the ratio of 15:21 or 5:7
then:
(1/2)axsinØ/( (1/2)bxsinØ ) = 5/7
canceling reduces this to
a/b = 5/7
5b = 7a
b = 7a/5 = 1.4a  ******
now let's use the perimeter:
a + b + 15+21 = 120
a+b =84 ******
sub in b = 1.4a into the above equation
a+b = 84
a + 1.4a = 84
2.4a = 84
a = 84/2.4 = 35
then b = 1.4(35) = 49
great question !!!!

Geometry8th gr  Steve, Friday, February 1, 2013 at 3:31pm
or, using the angle bisector theorem,
AD/DC = AB/BC
so, as Reiny calculated,
AB/BC = 5/7
AD+BC = 84
5x+12x=84
x=7
so, AB = 35
BC = 49

Geometry8th gr  typo  Steve, Friday, February 1, 2013 at 3:32pm
oops
5x+7x=84
x=7

Geometry8th gr  Mona, Friday, February 1, 2013 at 5:21pm
Thanks.
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