Unit 2, Lesson 15
Congruence for Quadrilaterals
Geometry
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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 \(EFGH\) is a rhombus, the distance from \(E\) to \(\underline{\hspace{.5in}1\hspace{.5in}}\) is the same as the distance from \(E\) to \(\underline{\hspace{.5in}2\hspace{.5in}}\). Since \(E\) is the same distance from \(\underline{\hspace{.5in}3\hspace{.5in}}\) as it is from \(\underline{\hspace{.5in}4\hspace{.5in}}\), it must lie on the perpendicular bisector of segment \(\underline{\hspace{.5in}5\hspace{.5in}}\). By the same reasoning, \(G\) must lie on the perpendicular bisector of \(\underline{\hspace{.5in}6\hspace{.5in}}\). Therefore, line \(\underline{\hspace{.5in}7\hspace{.5in}}\) is the perpendicular bisector of segment \(FH\). So reflecting rhombus \(EFGH\) across line \(\underline{\hspace{.5in}8\hspace{.5in}}\) will take \(E\) to \(\underline{\hspace{.5in}9\hspace{.5in}}\) and \(G\) to \(\underline{\hspace{.5in}10\hspace{.5in}}\) (because \(E\) and \(G\) are on the line of reflection) and \(F\) to \(\underline{\hspace{.5in}11\hspace{.5in}}\)and \(H\) to \(\underline{\hspace{.5in}12\hspace{.5in}}\) (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 \(EFGH\).
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 \(AD\) is parallel to \(\underline{\hspace{.5in}1\hspace{.5in}}\), alternate interior angles \(\underline{\hspace{.5in}2\hspace{.5in}}\) and \(\underline{\hspace{.5in}3\hspace{.5in}}\) are congruent. \(AC\) is congruent to \(\underline{\hspace{.5in}4\hspace{.5in}}\) since segments are congruent to themselves. Along with the given information that \(AD\) is congruent to \(BC\), triangle \(ADC\) is congruent to \(\underline{\hspace{.5in}5\hspace{.5in}}\) by the \(\underline{\hspace{.5in}6\hspace{.5in}}\) Triangle Congruence Theorem. Since the triangles are congruent, all pairs of corresponding angles are congruent, so angle \(DCA\) is congruent to \(\underline{\hspace{.5in}7\hspace{.5in}}\). Since those alternate interior angles are congruent, \(AB\) must be parallel to \(\underline{\hspace{.5in}8\hspace{.5in}}\). Since we define a parallelogram as a quadrilateral with both pairs of opposite sides parallel, \(ABCD\) is a parallelogram.
\(EFGH\) is a parallelogram, and angle \(HEF\) is a right angle. Select all statements that must be true.
\(EFGH \) is a rectangle.
Triangle \(HEF\) is congruent to triangle \(GFH\).
Triangle \( HEF\) is congruent to triangle \(FGH\).
\(ED \) is congruent to \(HD\), \(DG\), and \(DF\).
Triangle \(EDH\) is congruent to triangle \(HDG\).
Figure \(ABCD\) is a parallelogram. Is triangle \(ADB\) congruent to triangle \(CBD\)? Show or explain your reasoning.
\(\overline{AB} \cong \overline{CD}, \angle ADB \cong \angle CBD\)
Figure \(KLMN\) is a parallelogram. Prove that triangle \(KNL\) is congruent to triangle \(MLN\).