Geometry : 115

 Triangular area

In triangle ABC, produce a line from B to AC, meeting at D, and from C to AB, meeting at E.  Let BD and CE meet at X.
Let BXE have area a, BXC have area b, and CXD have area c.  Find the area of quadrilateral AEXD in terms of a, b, and c.

solution :   

BXE has area a, BXC has area b, and CXD has area c.

We will use the fact that the area of a triangle is equal to ½ × base × perpendicular height.  Any side can serve as the base, and then the perpendicular height extends from the vertex opposite the base to meet the base (or an extension of it) at right angles.

Consider BXE and BXC, with bases EX and XC, respectively.
The triangles have common height; therefore EX/XC = a/b.
Similarly, considering BXC and CXD, with respective bases BX and XD, BX/XD = b/c.

Now draw line AX.  Let AXE have area x and AXD have area y.

Consider AXB and AXD, with bases BX and XD, such that BX/XD = b/c.
Since AXB and AXD have common height, we have (a+x)/y = b/c.
Similarly, considering AXE and AXC, with bases EX and XC, x/(y+c) = a/b.

Cross-multiplying yields: by = cx + ac, bx = ay + ac.
Solving these eqns , we obtain x = ac(a + b)/(b² - ac), y = ac(b + c)/(b² - ac).

Therefore the area of quadrilateral AEXD =