In a triangle ABC, D and E are points on AB, AC respectively such that DE is parallel to BC. Suppose BE, CD intersect at O. If the areas of the triangles ADE and ODE are 3 and 1 respectively, find the area of the triangle ABC, with justification

Explanation

We denote the area of triangle PQR by [PQR]. We see that [BOD] and [COE] are equal. Let the common value be x, and let [BOC] = t. Using the fact that the ratio of areas of two triangles having equal altitudes is the same as the ratio of their respective bases, we obtain.
`(x)/(1)=(BO)/(OE)=(t)/(x)`.

This gives `t=x^(2)`. Now ADE and ABC are similar so that
`([ADE])/([ABC])=(DE^(2))/(BC^(2))=([ODE])/([OBC])` ,
since ODE and OCB are also similar. This implies that
`(3)/(4+2x+t)=(1)/(t)`,
which simplifies to t=2+x, using `t=x^(2)` we get a quadratic in ` x : x^(2)-x-2=0`. Its solution are x=2 and x=-1. Since x cannot be negative, x=2 and t=4. Thus `[ABC]=4+2x+t=4+4+4=12`