11524

# 22. Construct a Triangle Given the Hypotenuse and the Length of an Angle Bisector

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.
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. ■
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, .

### DETAILS

This Demonstration gives an alternative construction of Example 2 in [1, pp. 307, 308].
References
[1] B. I. Argunov and M. B. Balk, Elementary Geometry (in Russian), Moscow: Prosveščenie, 1966.
[2] G. E. Martin, Geometric Constructions, New York: Springer, 1998.

### PERMANENT CITATION

 Share: Embed Interactive Demonstration New! Just copy and paste this snippet of JavaScript code into your website or blog to put the live Demonstration on your site. More details » Download Demonstration as CDF » Download Author Code »(preview ») Files require Wolfram CDF Player or Mathematica.

#### Related Topics

 RELATED RESOURCES
 The #1 tool for creating Demonstrations and anything technical. Explore anything with the first computational knowledge engine. The web's most extensive mathematics resource. An app for every course—right in the palm of your hand. Read our views on math,science, and technology. The format that makes Demonstrations (and any information) easy to share and interact with. Programs & resources for educators, schools & students. Join the initiative for modernizing math education. Walk through homework problems one step at a time, with hints to help along the way. Unlimited random practice problems and answers with built-in step-by-step solutions. Practice online or make a printable study sheet. Knowledge-based programming for everyone.