Triangle areas

[sql_lall]

Ok, think of a triangle ABC
Lines come from B and C meeting AC and AB at D and E
now, the triangle is split up into 4 areas- 3 triangles and a quadrilateral.
I was wondering, is there some relationship between the areas of the triangles and that of the quad??

[Jared]

are the lines that intersect the triangles ACB and ABC bisectors, or just any random angle?

[sql_lall]

the lines BD and CE aren't anything special, they can be median, bisectors, altitudes, anthing.
Basically, they are just general Cevians

[riis]

This answer may sound stupid but it's a pretty obvious relationship.
When the areas of the triangles are smaller, then the area of the quadrilateral will be larger. The sum of all areas will be equal to the area of the large triangle.

Or do you want formula's of the relationship?

[prog_tom]

Draw a circumcircle, with circumradius of 1.

Suppose one of the sides of the quadralaterial is 1, then the relationship between the area of the triangle and the length of the circumcircle can be shown as:

[ABC] = abc/4R

where R is the circumradius.

Thus the quad' area can be solved from there.

[sql_lall]

prog_tom I'm not exatcly sure what you mean. Is this dependant on the circumradius and a side of the quad both being 1?

Riis Yeah, i know what u mean, however u r right, i'm looking for some way to write the area of the quad in terms of only the areas of the three triangles. The whole area is not given.

[prog_tom]

i meant a box. Box is a special case of quad.

[bugzpodder]

call the intersection F. [AEF]:[EFB]=[AEC]:[ECB]=AE:EB [XYZ] is area of XYZ. same relationship holds for other triangles.

[sql_lall]

Hey, dw, i figured out the answer
If the triangles are a, b, and c (b=one with side along BC)
Then area of quad=
(a + 2b + c)/(b²-ac)