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SQUARE construction from isosceles right triangle.
RECTANGLE, RHOMBUS, PARALLELOGRAM and TRAPEZOID construction.
The pictures show the constructions that prove the conjectures we’re about to show.
The top figure shows the right triangle AOB in bold blue outline. By placing three copies of this triangle together (dotted line triangles) with the original, square ABCD is formed. The conjectures are:
The diagonals have equal length.
They bisect one another at right angles.
The proof can be seen in the way the square has been constructed from four congruent triangles.
Now we come to the lower figure. We start with the red right triangle AOB where AO has length a and OB=b. The dotted magenta triangle has been placed so as to form the rectangle OAA'B. Four isosceles triangles are formed within the rectangle with X as their common vertex. From this construction we can prove the conjectures:
The diagonals are the same length and they bisect one another.
X is the midpoint of each diagonal.
Triangle AOB is copied and used to make the rhombus ABCD (dotted green triangles) to prove the conjecture that the diagonals bisect one another at right angles. The side length of the rhombus is √a²+b².
ODAA' is a parallelogram with diagonals OA and DA'. Triangles AYD and OYA are congruent. This proves the conjecture that the diagonals bisect one another at Y.
DAA'E is a trapezoid and we can work out the lengths of the diagonals and they are not equal.