CONSTRUCTION:
Construct a point F such that line FD is perpendicular to line AC
and |FD| = |AC|.
Construct a line m through D parallel to line BF.
Let P and S be the respective projections of A and C on line m.
Let Q and R be the respective projections of A and C on line BF.
PROOF:
Clearly, by construction, PQRS is a rectangle meeting the criteria of the problem. All we need to do is prove that |PQ| = |RQ| to show that PQRS is a square.
Let K be a point on line RS such that QK is parallel to line AC.
Let L be a point on line SP such that QL is parallel to line FD.
Clearly, QKCA and QLDF are parallelograms. Thus,
|QK| = |AC| = |FD| = |QL|
and
QK parallel AC perpendicular FD parallel QL
In the following, directed angles are used:
/LQP = /LQK - /PQK
= /PQR - /PQK
= /KQR
Therefore, LPQ and KRQ are congruent right triangles
and |PQ| = |RQ|.
COMMENTS:
If BD is not perpendicular to AC, then
Two solutions from above construction.
else
If |BD| = |AC|, then
Infinite number of solutions.
Construct parallel lines through A and C and
lines perpendicular to those through B and D.
else
No solutions.
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