Question

1. complete the statements below.

a kite is a quadrilateral with at least one pair of consecutive congruent sides.
the diagonals of a kite are perpendicular.
at least one diagonal of a kite bisects the other.

a trapezoid is a quadrilateral with at least one pair of opposite parallel sides.
the length of a trapezoid’s midsegment is the average of the lengths of the parallel sides.

an isosceles trapezoid is a trapezoid with congruent base angles. the diagonals of an isosceles trapezoid are congruent.

kite diagram with vertices a, b, c, d
if \\(\overline{ab} \cong \underline{\quad\quad}\\) and \\(\overline{bc} \cong \underline{\quad\quad}\\),
then \\(\overline{bd} \perp \underline{\quad\quad}\\) and \\(\overline{ac}\\) bisects \\(\underline{\quad\quad}\\).

trapezoid diagram with vertices e, f, g, d and midsegment xy
if \\(\overline{ef} \parallel \underline{\quad\quad}\\),
then \\(\underline{\quad\quad} = (ef + dg) \div 2\\).

isosceles trapezoid diagram with vertices j, k, l, h
if \\(\overline{jk} \parallel \underline{\quad\quad}\\) and \\(\underline{\quad\quad} \cong \overline{jh}\\),
then \\(\underline{\quad\quad} \cong \angle l\\) and \\(\overline{jl} \cong \underline{\quad\quad}\\).

Explanation:

Step1: Analyze the kite diagram (left)

In a kite, consecutive sides are congruent. So \( \overline{AB} \cong \overline{AD} \) (since \( AB \) and \( AD \) have the same tick marks) and \( \overline{BC} \cong \overline{DC} \) (same tick marks). The diagonals of a kite are perpendicular, so \( \overline{BD} \perp \overline{AC} \). Also, one diagonal bisects the other angles, so \( \overline{AC} \) bisects \( \angle BAD \) and \( \angle BCD \) (or \( \angle ABC \) and \( \angle ADC \), but from the diagram, \( \overline{AC} \) bisects \( \angle BAD \) and \( \angle BCD \); here, focusing on the angles at \( B \) and \( D \), \( \overline{AC} \) bisects \( \angle BAD \) and \( \angle BCD \), but the blanks: \( \overline{AB} \cong \overline{AD} \), \( \overline{BC} \cong \overline{DC} \), \( \overline{BD} \perp \overline{AC} \), \( \overline{AC} \) bisects \( \angle BAD \) (or \( \angle BCD \), but from the diagram, the angles at \( B \) and \( D \) are marked, so \( \overline{AC} \) bisects \( \angle BAD \) and \( \angle BCD \); for the blanks, \( \overline{AB} \cong \overline{AD} \), \( \overline{BC} \cong \overline{DC} \), \( \overline{BD} \perp \overline{AC} \), \( \overline{AC} \) bisects \( \angle BAD \) (or \( \angle BCD \), but the diagram shows \( \angle ABC \) and \( \angle ADC \) with marks, so \( \overline{AC} \) bisects \( \angle BAD \) and \( \angle BCD \); but the blanks: \( \overline{AB} \cong \overline{AD} \), \( \overline{BC} \cong \overline{DC} \), \( \overline{BD} \perp \overline{AC} \), \( \overline{AC} \) bisects \( \angle BAD \) (or \( \angle BCD \); here, the first two blanks: \( \overline{AD} \), \( \overline{DC} \); then \( \overline{AC} \); then \( \angle BAD \) (or \( \angle BCD \), but from the diagram, the angles at \( B \) and \( D \) are marked, so \( \overline{AC} \) bisects \( \angle BAD \) and \( \angle BCD \); so \( \overline{AB} \cong \overline{AD} \), \( \overline{BC} \cong \overline{DC} \), \( \overline{BD} \perp \overline{AC} \), \( \overline{AC} \) bisects \( \angle BAD \) (or \( \angle BCD \)).

Step2: Analyze the trapezoid diagram (middle)

A trapezoid has at least one pair of parallel sides. The midsegment (or median) of a trapezoid is the average of the two parallel sides. So if \( \overline{EF} \parallel \overline{DG} \), then the midsegment \( \overline{XY} \) has length \( \frac{EF + DG}{2} \). So \( \overline{EF} \parallel \overline{DG} \), then \( XY = \frac{EF + DG}{2} \).

Step3: Analyze the isosceles trapezoid diagram (right)

An isosceles trapezoid has \( JK \parallel \overline{HL} \) (since \( JK \) and \( HL \) are the two bases with arrows indicating parallelism). The legs \( JH \) and \( KL \) are congruent (so \( \overline{KL} \cong \overline{JH} \)). The base angles are congruent, so \( \angle H \cong \angle L \) (wait, the diagram has \( \angle H \) and \( \angle L \) marked, but the blank is \( \_\_\_ \cong \angle L \), so \( \angle H \cong \angle L \)? Wait, no, in isosceles trapezoid, base angles are congruent. So \( JK \parallel \overline{HL} \), \( \overline{KL} \cong \overline{JH} \), then \( \angle H \cong \angle L \) (or \( \angle J \cong \angle K \)), and diagonals are congruent, so \( \overline{JL} \cong \overline{KH} \).

Answer:

(for each blank):

• Kite (left): \( \overline{AB} \cong \boldsymbol{\overline{AD}} \), \( \overline{BC} \cong \boldsymbol{\overline{DC}} \), \( \overline{BD} \perp \boldsymbol{\overline{AC}} \), \( \overline{AC} \) bisects \( \boldsymbol{\angle BAD} \) (or \( \angle BCD \))
• Trapezoid (middle): \( \overline{EF} \parallel \boldsymbol{\overline{DG}} \), then \( \boldsymbol{XY} = \frac{EF + DG}{2} \)
• Isosceles Trapezoid (right): \( \overline{JK} \parallel \boldsymbol{\overline{HL}} \), \( \boldsymbol{\overline{KL}} \cong \overline{JH} \), then \( \boldsymbol{\angle H} \cong \angle L \), \( \overline{JL} \cong \boldsymbol{\overline{KH}} \)

(Note: The exact blanks depend on the problem's formatting, but the key is using the properties of kite, trapezoid, and isosceles trapezoid.)