CEVAŐS THEOREM

BY

SHADRECK S CHITSONGA

 

PROOFS

Here we will look at the proofs to the conjectures we made earlier on. We will look at three situations.

 

1. When P is the centroid of triangle ABC

Given:          with P as the centroid of the triangle, such

                that    and  , produced meet the sides of

                 at E, F and D respectively.

To prove:     that

Proof:         P is the centroid (given)

              D,E, and F are the mid-points of sides BC,CA, and

                 AB respectively.

                =, = , and = . (i)

                 Now let us consider the products (AF)(BD)(CE) and       

                 (BF)(CD)(AE).

                 Using (i) we see that the products are equal to each 

                  other.

              , holds.

 

2. When P is the orthocenter of the triangle

 

 

Given:          with P as the orthocenter of the triangle ,

                such that     and  , produced meet the

                sides of  at E,F and D  respectively.                   

To prove :    that

Proof:          P is the orthocenter (given)

                 A line from the vertex of a triangle that passes

                 through the orthocenter is perpendicular to the  

                 opposite side.

            The angles AFP, BFP, BDP, PDC, CEP, PEA are all   

               equal to 90ˇ , therefore equal to each other.   

               In triangles BPF and CPE,

                = 90ˇ (proved).

                (vertically opposite angles).

             (angle sum of triangle =180ˇ)

              is similar to .

               (i)

                Similarly we can show that  is similar to  

                .

                  (ii)

                    Also  is similar to .

                  (iii),

             From (i)  and from (iii) 

            

                  (iv)

                    From (iii) , and we can substitute

                    this in (iv).

                  

                    Dividing both sides by AP and some      

                    rearrangements we end up 

                    with:   (v)

                    You will notice that (v) is equivalent to

                   

 

3. When P is the incenter

 

Given:         with P as the incenter  triangle ,

               such that     and  , produced meet the

                sides of   at E,F and D  respectively. 

To prove:     that

Proof :        Before doing this proof we need some background    

                information. We need to make use of another 

                theorem. This theorem states that:

 

The bisector of an interior angle of a triangle divides the opposite side into segments proportional to the adjacent sides. Look at the GSP in figure 1.

Figure 1.

 

NOTE: This is not a proof for this theorem. We are just using the result here.

Using the results in figure 1, we now make use of the ratios

(BD/DC)=(AB/AC), (AC/BC)=(AF/FB), and (AE/EC)=(AB/BC).

We can rearrange these ratios as

(i) AB={(BD X AC)/DC)}  (ii) AB = {(AE X BC)/ CE)} 

(iii) AC = {(FA X BC)/BF}

Using (i) and (ii) we have {(BD X AC)/DC)}= {(AE X BC)/ CE)} (iv).

Now substituting (iii) in (iv) we have:

{(BD )(FA X BC)/(BF X DC)} = {(AE X BC)/(CE)}

this simplifies to (BD) (FA) (CE) = (AE) (BF) (DC).

Therefore {(BD) (FA) (CE)}/ {(AE) (BF) (DC)} = 1.

 

 

4. When P is an arbitrary point inside the triangle.

 

 

Given:              with P as an arbitrary point inside the triangle ,

                     such that     and  , produced meet the sides of

                       at E,F and D  respectively.

To prove:          that

Construction:      Draw  parallel to , to meet  in Y.

                     Draw  parallel to  to meet  in X.

Proof:               is parallel to  (construction) .

                    also      

                     We can write  also  

                   (i)

                    Equation (i) can also be expressed as

        is  parallel to  (construction)

        also

      We can write   also 

     (ii)

Equation(ii) can also be expressed as

                   Because equations (i) and (ii) are both equal to 1,

                   We can write

                   After simplifying we end up with:

 

      , which is equivalent to 

 

*NOTE: I considered including the converse after reading your      

               comments.

 To prove the converse, we must show that if  , holds then there must be a a common point P where all the three Cevians pass through.

GIVEN:              , where D, E, and F are points on the sides         

                       BC, AC, and AB  respectively.

 TO PROVE:         That segments BE, AD and CF meet a common point( say P)

PROOF:              , , and  are the sides of triangle ABC, therefore they

                       are all not parallel to each   other.

                       We will assume that the segments BE, AD, and CF are not 

                       parallel to each other.

                       With this assumption, then two of the segments must intersect

                        at a point.

                        Let us consider segments BE and CF and let them intersect     

                        in P. This means that the two segments contain the point P.

                        Now we can draw a line AP. If we assume that that line AP is

                        not parallel to segment BC, then the two must intersect at

                         point. Let us call this point DŐ on the segment BC.

                         If this is the case then we have segments BE, CF and ADŐ

                         passing through a common point P.

                         This then means that , must hold.

                         But we have been given that .

                         We can see that , which means that DŐ and D are    

                         the same point.

                         We can therefore conclude that all the three segments AD,

                         BE, and CF contain the same point P, therefore they are

                         concurrent.