A homework question I have been having some issue with -
Given parallelogram $ABCD$, generate 4 squares from the sides of the $ABCD$. Given the 4 centers of the squares $W, X, Y, Z$ (formed by their respective squares defined by sides $AB, BC, CD, DA$ respectively), show the centers form a new square.
The first step would be to show two sides, say $ZW$ and $WX$, are equal. This can be done by similar triangles. $AW$ and $WB$ are equal because they are both half the length of the diagonal of the same square. $BX$ and $AZ$ are equal because of the symmetry line $DB$ (flip the parallelogram over and its equal).
Which leads me to where Im stuck at. I need to show 2 more things to finish the proof.
- $\angle WBZ = \angle ZAW$. This completes the side-angle-side, proving the triangles are congruent and their third side is equal => the centers form at least a rhombus. I believe this is done using interior/exterior angles, but I dont see how.
- $WY \perp ZX$. This brings the rhombus to a square, thus completing the proof. I dont have a clue how to do this.