We have successfully proved that in rhombus WXYZ, WY and XZ are perpendicular bisectors of each other.
Theorem 6-16: In rhombus WXYZ, WY and XZ are perpendicular bisectors of each other.
Proof:
Given: Rhombus WXYZ
To prove: WY and XZ are perpendicular bisectors of each other.
Proof Steps:
1. Show that WXYZ is a rhombus:
A rhombus is a quadrilateral with all sides of equal length. Given that WXYZ is a rhombus, it means that all sides are equal.
2. Show that opposite sides of the rhombus are parallel:
In a rhombus, opposite sides are parallel. This is a property of rhombi.
3. Show that opposite angles of the rhombus are equal:
In a rhombus, opposite angles are equal. This is another property of rhombi.
4. Show that diagonals bisect each other:
Diagonals of a rhombus bisect each other at right angles. This is a well-known property of rhombi.
5. Show that WY and XZ are perpendicular:
From step 4, we know that the diagonals bisect each other at right angles. Therefore, WY and XZ are perpendicular.
6. Show that WY and XZ are bisectors:
From step 4, we know that the diagonals bisect each other. Therefore, WY and XZ are bisectors.
7. Conclude:
WY and XZ are perpendicular bisectors of each other. This is because they are perpendicular (step 5) and bisect each other (step 6).
Hence proved.