Question 389081
Instead of Pythagorean-theorem bashing, I'm only going to find the change in x and y for each pair of points:


(The partial differential operator {{{delta}}} should be uppercase, which represents a total change. Also, I'm taking the absolute value since Pythagorean theorem only depends on the squares of the numbers, regardless of whether change is positive or negative.


A,B
{{{delta(x) = 3}}}
{{{delta(y) = 4}}}


B,C
{{{delta(x) = 4}}}
{{{delta(y) = 1}}}


C,A
{{{delta(x) = 1}}}
{{{delta(y) = 3}}}


D,E
{{{delta(x) = 4}}}
{{{delta(y) = 3}}}


E,F
{{{delta(x) = 1}}}
{{{delta(x) = 4}}}


D,F
{{{delta(x) = 3}}}
{{{delta(x) = 1}}}


We see that AB = DE = 5, AC = DF = {{{sqrt(10)}}}, BC = EF = {{{sqrt(17)}}}. By SSS, ABC and DEF are congruent triangles.