Quadrilaterals

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Two proofs.

Adapted from Ohio’s 2017 Geometry released item 13.

Two pairs of parallel lines intersect to form a parallelogram as shown.

Complete the following proof that opposite angles of a parallelogram are congruent:

(a): $\angle 1 \cong \angle 2$ as opposite anglesalternate interior anglescorresponding angles for parallel lines $m$ and $n$ $k$ and $l$ .
(b): $\angle 3 \cong \angle 2$ as opposite anglesalternate interior anglescorresponding angles for parallel lines $m$ and $n$ $k$ and $l$ .
(c): Then $\angle 1 \cong \angle 3$ because they are both congruent to $\angle 2$ .

Adapted from Ohio’s 2018 Geometry released item 21.

Given the parallelogram $WXYZ$ , prove that $\overline {WX}\cong \overline {YZ}$ .

Fix note: It really would help to have an online environment that allows students to mark diagrams.

Complete the proof below:

(a): $\angle ZWY \cong \angle XYW$ as alternate interior anglescorresponding anglesopposite angles for parallel segments $\overline {WZ}$ and $\overline {XY}$ $\overline {WX}$ and $\overline {YZ}$ .
(b): $\angle ZYW \cong \angle XWY$ for the same reason, this time for parallel segments $\overline {WZ}$ and $\overline {XY}$ $\overline {WX}$ and $\overline {YZ}$ .
(c): $\overline {WY}\cong \overline {YW}$ because a segment is congruent to itself.
(d): $\triangle WYZ \cong \triangle YWX$ by SASASASSS .
(e): Then $\overline {YZ}\cong \overline {WX}$ as corresponding parts of congruent triangles.

Fix note: Maybe number the angles.

Quadrilateral $ABCD$ is a kite as marked. Prove that $\overleftrightarrow {BD}$ is the perpendicular bisector of $\overline {AC}$ .

Key theorem: The points on a perpendicular bisector are exactly those that are equidistant from the endpoints of a segment.

Proof: Because $B$ and $D$ are each $\answer [format=string]{equidistant}$ from $A$ and $C$ , they each must lie on the perpendicular bisector of segment $\answer {AC}$ , which implies that $\overleftrightarrow {BD}$ is its perpendicular bisector.

Quadrilateral $ABCD$ is a kite as marked. Prove that $\overleftrightarrow {BD}$ is the perpendicular bisector of $\overline {AC}$ .

A proof that makes use of triangle congruence:

Fix note: Do we need a step about $\overleftrightarrow {BX}$ and $\overleftrightarrow {BD}$ being the same line?

In the proof above, $\triangle BAD \cong \triangle BCD$ by $\answer [format=string]{SSS}$ , and $\triangle ABX \cong \triangle CBX$ by $\answer [format=string]{SAS}$ .