Question

move vertex d and observe how the measures change. make a conjecture. which of the following statements appear to be true? check all that apply. ac and bd are always equal. ae and ce are always equal. be and de are always equal. ac and bd always bisect each other. ac and bd are always perpendicular. diagram: quadrilateral abcd with diagonals intersecting at e. ae = 40 units, ce = 40 units, be = 32 units, de = 32 units

Explanation:

• For "AE and CE are always equal": From the given data, \( AE = 40 \) and \( CE = 40 \), and moving vertex \( D \) (in a parallelogram - like figure, as diagonals bisect each other) keeps \( AE = CE \) (diagonals of a parallelogram bisect each other, so \( E \) is mid - point of \( AC \), hence \( AE = CE \)).
• For "BE and DE are always equal": Given \( BE = 32 \) and \( DE = 32 \), and in a parallelogram - like figure, \( E \) is mid - point of \( BD \) (diagonals bisect each other), so \( BE = DE \).
• For "AC and BD always bisect each other": The equal lengths of \( AE = CE \) and \( BE = DE \) imply that \( E \) is the mid - point of both \( AC \) and \( BD \), so diagonals bisect each other.
• "AC and BD are always equal": In a parallelogram, diagonals are not always equal (they are equal in rectangles, a special case of parallelogram). From the given lengths (\( AC=AE + CE=80\), \( BD = BE+DE = 64\)), they are not equal, so this is false.
• "AC and BD are always perpendicular": In a general parallelogram, diagonals are not perpendicular (perpendicular in rhombus, a special case). So this is false.

Answer:

AE and CE are always equal, BE and DE are always equal, AC and BD always bisect each other.