This Demonstration constructs a right triangle

given the length

of the hypotenuse

and the length of the angle bisector

at

. Let

and

be the length of the legs.

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

.

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. ■

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

. ■

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,

.