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.

- Geometry-8th 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 !!!!

- Geometry-8th 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

- Geometry-8th gr - typo -
**Steve**, Friday, February 1, 2013 at 3:32pm
oops

5x+7x=84

x=7

- Geometry-8th gr -
**Mona**, Friday, February 1, 2013 at 5:21pm
Thanks.

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