Length of AD inside a triangle ABC

[songoku]

Homework Statement

Please see below

Relevant Equations

Cosine Rule

Area of triangle = 1/2 . a . b . sin C

I get the answer but my working is really long:
1) Find all the length of sides of the triangle
2) Let DB = x, so CD = CB - x
3) Compare the area of triangle ADC and ABD using formula 1/2 . a . b sin θ then find x
4) Find cosine of angle B by using cosine rule on triangle ABC
5) Use cosine rule again on triangle ABD to find the answer

Is there another approach to this question? Thanks

 

[Delta2]

I think steps 1-3 are kind of mandatory in order to find x.
But I "feel" there must be an easier way to find AD once you have found x. Hold on while I think a bit more on this.

 

[Delta2]

Only other thing I can think at the moment is to use Heron's formula for the triangle ABD. You know two sides and the area (1/4 of the area of the ABC) so you can find the third side.

 

[Ibix]

Because BD and BC are colinear and the triangles share the point A what does the area ratio tell you about the length ratio of BD and BC? Then work with the coordinates.

 

[songoku]

Thank you very much for the help Delta2 and Ibix

 

[Prof B]

Because BD and BC are colinear and the triangles share the point A what does the area ratio tell you about the length ratio of BD and BC? Then work with the coordinates.

The key thing is that ABC and ABD have the same altitude.