It is given that Y W bisects the angle of the parallelogram at vertex Y and at vertex W. This allows us to write the congruence of two pairs of angles of triangles △ Y Z W and △ Y X W.
∠ 3&≅ ∠ 4
∠ 1&≅ ∠ 2
We can see that Y W is a common side of the two triangles, so according to the Angle-Side-Angle (ASA) Congruence Postulate the two triangles are congruent.
△ Y Z W≅△ Y X W
Corresponding sides of congruent triangles are congruent.
Y Z&≅Y X
Z W&≅X W
This tells us that two pairs of adjacent sides of parallelogram WXYZ are congruent.
According to Theorem 6.3, opposite sides of parallelogram WXYZ are also congruent.
Y X&≅Z W
Together with the previous congruences this means that all four sides are congruent, so parallelogram WXYZ is a rhombus. We can summarize the process above in a paragraph proof.
Completed Proof
2
&Given:&& WXYZ is a parallelogram
& && WYbisects∠ ZWXand∠ ZYX
&Prove:&& WXYZ is a rhombus
Proof:
Since diagonal WY bisects the angles of the parallelogram at vertices W and Y, triangles △ YZW and △ YXW have two pairs of congruent angles. The included side, WY, is common in these two triangles, so △ YZW and △ YXW are congruent by ASA. Corresponding sides of congruent triangles are congruent, so YZ≅YX and ZW≅XW. Opposite sides of a parallelogram are also congruent, so YX≅ZW. Putting these congruences together, the transitive property of congruence implies that all four sides of parallelogram WXYZ are congruent, so WXYZ is a rhombus.
