Final Assignment

By

Cara Haskins

 

The Final Assignment will begin with any triangle ABC.  P is any point inside the triangle ABC, and EFD are the intersecting points of the segments drawn from the vertices of triangle ABC going through any point P.

The above illustrations demonstrate the relationships between line segments AF, BD, and CE in proportion to line segments FB, DC, and EA.  Notice that the size of the triangle may change, and the position of point P inside the triangle may change, and yet through it all the product of the orange line segments will always equal the product of the blue line segments.  We may then conclude from this that the ratio (AF)(BD)(CE)/(BF)(CD)(AE) will always equal 1.  You don’t believe me?  Click here to try for yourself. This is known as Ceva’s Theorem, which states that in a triangle ABC, three lines AD, BE, and CF intersect at a single point P if and only if (AF)(BD)(CE)/(FB)(DC)(EA)=1

 

Now this must be proved.

 

“Proofs are to mathematics what spelling (or even calligraphy) is to poetry. Mathematical works do consist of proofs, just as poems do consist of characters.”

 

Vladimir Arnold

 

Okay, so I did not come up with this completely on my own.  I must give credit to www.cut-the-knot.org/Generalization/ceva.shtml. 

 

First extend the line segments beyond the triangle until they meet the parallel line, which was made through point B parallel to AC.  Now we see several pairs of similar triangles: BJD and CAD, BFI and CAF, BIP and CEP, AEP and BJP.

 

 

From the similar triangles we conclude that

1.     BD/DC = BJ/CA

2.    AF/FB = CA/BI

3.     BI/CE = BP/EP

4.   BJ/EA = BP/EP

From the last two we conclude that

5.  BI/CE = BJ/EA and therefore,

6.  CE/EA = BI/BJ

Now, multiplying #1, #2, and #6 we see:

(BD/DC)(AF/FB)(CE/EA) = (BJ/CA)(CA/BI)(BI/BJ) = (BJ)(CA)(BI)/(CA)(BI)(BJ) = 1

 

Click here to see if it will always equal 1, with point P inside or outside of the Triangle ABC.

 

Okay, just in case you are looking for even more fun, let me show that when P is inside triangle ABC, the ratio of the areas of triangle ABC and triangle DEF is always greater than or equal to 4. 

Click here to move point P to different locations inside the triangle ABC and see what happens to the area of the ratio.  The ratio of the areas of triangle ABC and triangle DEF equal exactly four only when D, E, and F are the midpoints of triangle ABC and point P is the centroid of the triangle.

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