Answer:
Givens
[tex]DR=PT=3cm[/tex] and [tex]RP=4cm[/tex]
Points D, R, P and T are collinear.
The image attached shows a representation of this problem.
By sum of segments, we have
[tex]DT=DR+RP+PT[/tex]
Replacing all given values, we have
[tex]DT=3+4+3=10 \ cm[/tex]
Therefore, the segment DT is 10 centimeters long.
Now, we know that point E is the center of segment RP, which means is a middle point, divides the segment equally. So, to prove that segment DT is symmetrical regarding point E, we need to prove that both sides are equal.
[tex]RE=EP=2cm[/tex], by definition of middle point.
By sum of segments, we have
[tex]DR+RE=3+2=5cm\\EP+PT=5cm[/tex]
Which means [tex]DR+RE=\frac{DT}{2}\\ EP+PT=\frac{DT}{2}[/tex].
Therefore, segment DT is symmetrical regarding point E.
