Unit 1, Lesson 14
Congruence for Quadrilaterals
Integrated Math 2
Sign in to view assessments and invite other educators
Sign in using your existing Kendall Hunt account. If you don’t have one, create an educator account.
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 \(HDG\).
\(ABDE\) is an isosceles trapezoid. Priya makes a claim that triangle \( AEB\) is congruent to triangle \(DBE\). Convince Priya this is not true.
In quadrilateral \(ABCD\), triangle \(ADC\) is congruent to \(CBA\). Show that \(ABCD\) is a parallelogram.