# Lisa Brock

Final Assignment

Ceva’s Theorem

Consider any triangle ABC.  Select point P inside the triangle.  Draw lines from A, B, and C through P.  Let points D, E, and F be the intersections with the opposite side of the triangle.

Explore (AF)(BD)(CE) and (BF)(CD)(AE) for various triangles.  Click HERE to open a GSP sketch.  Move P around inside the triangle.  Also, move the vertices of the triangle around.

Notice that (AF)(BD)(CE) and (BF)(CD)(AE) are always equal.  Do you think this always going to be true when P is inside of the triangle?

, this will always be the case.  For a proof of Ceva’s Theorem, click HERE.

What if P is outside of the triangle?  Extend the segments joining A, B, and C into lines.  Move P outside of the triangle.  Click HERE to open a GSP sketch.

The two products are still equal!  The earlier proof supports this.  Click HERE to view the proof again.  Remark 2 of this proof addresses the situation when P is outside the triangle.

Now lets draw the triangle DEF.  Look at the ratio of the area of triangle ABC to triangle the area of triangle DEF when P is inside the triangle.  Click HERE to open a GSP sketch.

You should have noticed that the ratio of the areas is always greater than or equal to 4.  This can be proven using Routh’s Theorem.  This theorem gives a formula for the area of triangle DEF.  The formula is

where l, m, n are the ratios  ,respectively.

Rearranging this equation gives us a formula for the ratio of the areas of the two triangles.

We have already shown that lmn=1, so lmn+1=2.  The equation now looks like this:

(l+1)(m+1)(n+1) has a minimum value of 8.  This occurs when l,m,n=1.  When l,m,n=1, triangle DEF is the medial triangle.  Since (l+1)(m+1)(n+1) has a minimum value of 8, the ratio of the areas has a minimum value of 4.

In conclusion, when P is inside the triangle, the ratio of the area of triangle ABC to the area of triangle DEF is at least 4, with a minimum value of 4 occurring when triangle DEF is the medial triangle of triangle ABC.