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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