Given: CDEF is a parallelogram; S and T are the midpoints of EF and ED----- Prove:SR is congruent to FD?

Question

F S E
T
C D R

Answer 1

I think you mean:
.....F..S..E
............T
.C......D..R

And I think that the given is:
______________________________
GIVEN:
CDEF is a parallelogram
S is the midpt. of EF
T is the midpt. of ED
R is the intersection of lines ST and CD

PROVE:
SR = FD

Here are the steps:
1 .Since S is the midpt of EF .......... (given)

2 ....and T is the midpt of ED .......... (given)

3 .Then, EF = 2ES and ED = 2ET .. (definition of midpoint)

4 .Then, EF/ES = 2 and ED/ET = 2. (dividing)

5 .Then, EF/ES = ED/ET ................ (Transitive Property)

6Since, angle E = angle E, ............ (Reflexive Property)

7.Then, ΔSET ~ ΔFED .................. (SAS Similarity Theorem)

8.Thus, FD = 2ST .......................... (Definition of Similar Triangles)

9..Now, angle STE = angle RTD ... (Vertical Angle Theorem)

10....and ET = TD ............................ (Definition of Midpoint)

11.Since CDEF is a parallelogram . (given)

12.Then FE || CD ............................ (Definition of Parallelogram)

13then angle EST = angle DRT ...... (If a transversal cuts 2 || lines, then alternate int. angles are congruent)

14thus, ΔSET = triangle RDT .. (ASA Postulate)

15Then, ST = TR ..................... (Corresponding parts of congruent triangles are congruent)

16Thus, T is the midpt of SR. .. (Definition of Midpoint)

17Therefore, SR = 2ST ............ (definition of midpoint)

18Therefore, SR = FD ............. (Transitive Property of Equality, Statements 8 and 17)

Answer 2

List all the givens, which must also include something you left out, possibly SR = 2ST or ST congruent to SR

Then use that a side of a triangle (FD of triangle FDE) is twice as long as the segment connecting the midpoints of the other two sides (ST). Show by substitution that this makes it congruent to SR since it also is twice as long as ST. Kind of tough as this is just a guess on the missing given.

Answer 3

It's not possible for SR to be congruent to FD. Parallel, maybe, but you need to define R better.