Given the quadrilateral [math]ABCE[/math], the square [math]ABCD[/math], and the information that [math]F[/math] is the same distance from [math]A[/math] and [math]C[/math], show that [math]ABCE[/math] is symmetrical along [math]\overline{BE}[/math].
[list=1] [*]Recall the definition of line symmetry. [*]Since [math]\overline{AB} \cong \overline{BC}[/math] and [math]\overline{AF} \cong \overline{FC}[/math], [math]\overline{BF}[/math] is a line of symmetry for [math]\bigtriangleup{ABC}[/math] where [math]\bigtriangleup{ABF} \cong \bigtriangleup{CBF}[/math]. [*][math]\bigtriangleup{AFE}[/math] has the same area as [math]\bigtriangleup{CFE}[/math] because they share a base and have equal height. [math]\overline{AF} \cong \overline{FC}[/math], so [math]\bigtriangleup{AFE} \cong \bigtriangleup{CFE}[/math]. [*]We now know [math]\overline{FE}[/math] is a line of symmetry for [math]\bigtriangleup{ACE}[/math] and [math]\overline{BF}[/math] is a line of symmetry for [math]\bigtriangleup{ABC}[/math] , so [math]\bigtriangleup{ABE} \cong \bigtriangleup{CBE}[/math] and quadrilateral [math]ABCE[/math] is symmetrical along [math]\overline{BE}[/math]. [/list] This applet is provided by Walch Education as supplemental material for the [i]CCGPS Coordinate Algebra[/i] program. Visit [url="http://www.walch.com"]www.walch.com[/url] for more information on our resources.