SOLUTION: Can there exist a triangle ROS in which the trisectors of angle O intersect RS at D and E with RD =1, DE=2,and ES=4? Explain. Thanks to anyone who will help me with this! (P.S.

Make a sketch.  I will refer to it as if you have it in front of you (as I have it in front of my eyes).


    +---------------------------------------------------------------+
    |    Let ROS be a triangle, and let its angle ROS at vertex O   |
    |    is trisected in a way that RD= 1, DE=2 and ES=4.           |    (*)
    +---------------------------------------------------------------+


I am going to bring it to contradiction and prove, in this way, that such trisection IS NOT POSSIBLE.



Let "t" be any of the three congruent small angles at vertex O.

Let "a" be the length of RO;  let "b" be the length of DO.


Then  the length of  EO  is  2a  (bisector theorem for triangle REO), 
and   the length of  SO  is  2b  (bisector theorem for triangle DOS).


    I will remind you, that the bisector theorem STATES that
    the ratio of the segments lengths, to which the bisector of a triangle
    divides the opposite side is equal to the ratio of the lengths
    of the two lateral corresponding sides.
    
    
Now,  triangle ROD has side lengths "a" and "b", that conclude the angle "t", and the opposite side RD= 1.

So,  we write the Cosine Law formula for RD

     RD^2 =  = 1^2.                  (1)



Next, triangle DOE has side lengths "b" and "2a", that conclude the angle "t", and the opposite side DE= 2.

So, we write the Cosine Law formula for DE

    DE^2 =  = 2^2.            (2)



Finally, triangle EOS has side lengths "2a" and "2b", that conclude the angle "t", and the opposite side ES= 4.

So, we write the Cosine Law formula for ES

    ES^2 =  = 4^2.      (3)



I will rewrite the three equations (1), (2) (3) in simpler form

             = 1.                   (1')

             = 4.                  (2')

             = 16.                (3')


Now compare equations  (1')  and  (3').   You see that the left side of (3') is four times the left side of (1'),

but the right side of (3') is sixteen times left side of (1').


It is the contradiction, and this contradiction proves that the original assumption (*) is not valid / (is not possible).

    Indeed, from one hand side, triangles ROD and EOS must be similar with the similarity coefficient of 2,

    because their "lateral" sides are proportional  RO : DO = a : b = EO : SO = (2a) : (2b) and the concluded angles "t" are congruent;


    but from the other hand side,  their sides, opposite to angle "t", are with the ratio 4:1, which gives the CONTRADICTION.