IM Geo.2.15 Practice: Congruence for Quadrilaterals
Select all quadrilaterals that have 180 degree rotational symmetry.
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.
Because is a rhombus, the distance from to 
is the same as the distance from to 
. Since is the same distance from 
as it is from 
, it must lie on the perpendicular bisector of segment 
. By the same reasoning, must lie on the perpendicular bisector of 
. Therefore, line 
is the perpendicular bisector of segment . So reflecting rhombus across line 
will take to 
and to 
(because and are on the line of reflection) and to
and to 
(since is perpendicular to the line of reflection, and and are the same distance from the line of reflection, on opposite sides). Since the image of rhombus reflected across is rhombus (the same rhombus!), line must be a line of symmetry for rhombus .
In quadrilateral ABCD, AD is congruent to BC, and AD is parallel to BC.
Andre has written a proof to show that ABCD is a parallelogram.
Fill in the blanks to complete the proof.
Since is parallel to 
, alternate interior angles 
and 
are congruent. is congruent to 
since segments are congruent to themselves. Along with the given information that is congruent to , triangle is congruent to
by the 
Triangle Congruence. Since the triangles are congruent, all pairs of corresponding angles are congruent, so angle is congruent to 
. Since those alternate interior angles are congruent, must be parallel to 
. Since we define a parallelogram as a quadrilateral with both pairs of opposite sides parallel, is a parallelogram.
Select the statement that must be true.
EFGH is a parallelogram and angle HEF is a right angle.
Select all statements that must be true.
Figure ABCD is a parallelogram.
Is triangle congruent to triangle ? Show or explain your reasoning.
Figure KLMN is a parallelogram.
Prove that triangle is congruent to triangle .