given AF=PC, DN = FC prove DN = AP

Question

given AF=PC, DN = FC prove DN = AP

Answer 1

is AFCP a rectangle or parallelogram ?

if so, FC=AP because opposite sides of a parallelogram are congruent

DN=FC is given

if DN=FC and FC=AP, then DN=AP by the transitive property of equality

Answer 2

If A, F, P, C, D, and N are points on the number line where the length is the second coordinate minus the first, this becomes algebra.

AF = PC, so F - A = C - P (Equation 1)

DN = FC, so N - D = C - F (Equation 2)

if F - A = C - P, adding P - F to both sides,

F - A + P - F = C - P + P - F,

P - A = C - F (Equation 3)

Then, if C - F = N - D (Equation 2), and C - F = P - A (Equation 3), from the transitive law, N - D = P - A

Writing this as a line segment, DN = AP