Construction

Step 1. Let have length .

Let the point be on the perpendicular to at such that . Let be the circle with center and radius .

Step 2. Let and be the intersections of the ray and .

Let the points and be on so that and . Let be the point on the extension of so that is the midpoint of .

Step 3. Let be the intersection of the ray and the perpendicular to at .

Step 4. Let be the midpoint of . Let be the circle with center and radius . Let be the intersection of and the parallel to through .

Step 5. Draw the triangle .

Verification

From step 1, .

Step 4 and Thales's theorem imply that triangle is right-angled at .

It remains to prove that the length of the angle bisector at is .

Theorem 1. Let be a point on the hypotenuse , and let and be the perpendicular projections of on and , respectively. Then the rectangle is a square if and only if is the angle bisector at .

Proof. If is a square, then the diagonal forms angles with and , so it bisects the angle at .
If is the angle bisector at , then and are equilateral right-angled congruent triangles and therefore . So is a square. \[FilledSquare]

Theorem 2. Let be a right triangle with hypotenuse , legs and , and where the length of the angle bisector is . Then is a solution of the equation , and the altitude is given by .

Proof. Let be the angle bisector at with on and let be as described in Theorem 1. The side length of the square is , since is the length of its diagonal. We can express twice the area of the triangle as . Since , we have a quadratic equation for , namely, or .
Twice the area of is so . \[FilledSquare]

The first equation of theorem 2 gives a construction of using the power of with respect to . In this case, by step 2, , since .

The second equation of theorem 2 gives a construction of using similarity of triangles. Namely, by step 3, is such that is similar to . So from .

Since is a right triangle with hypotenuse and altitude , using theorem 2, .

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