Lin wrote a proof to show that diagonal EG is a line of symmetry for rhombus
EFGH. Fill in the blanks to complete her
proof
G
Because EFGH is a rhombus, the distance from Eto F 1 is the same
as the distance from Eto H 2 Since E is the same distance from
3 as it is from 4 , it must lie on the perpendicular
bisector of segment 5 By the same reasoning, G must lie on the
perpendicular bisector of 6 Therefore, line 7
is the
perpendicular bisector of segment FH. So reflecting rhombus EFGH across line
8 will take Eto 9 and G to 10
(because E and G are on the line of reflection) and F to 11 and H to
12 (since FH is perpendicular to the line of reflection, and F and H
are the same distance from the line of reflection, on opposite sides). Since the image
of rhombus EFGH reflected across EG is rhombus EHGF (the same rhombus!),
line EG must be a line of symmetry for rhombus EFGH.
(From Unit 2, Lesson 14.)