11453

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, .

SNAPSHOTS

  • [Snapshot]
  • [Snapshot]
  • [Snapshot]

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.
    • 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 »

Files require Wolfram CDF Player or Mathematica.









 
RELATED RESOURCES
Mathematica »
The #1 tool for creating Demonstrations
and anything technical.
Wolfram|Alpha »
Explore anything with the first
computational knowledge engine.
MathWorld »
The web's most extensive
mathematics resource.
Course Assistant Apps »
An app for every course—
right in the palm of your hand.
Wolfram Blog »
Read our views on math,
science, and technology.
Computable Document Format »
The format that makes Demonstrations
(and any information) easy to share and
interact with.
STEM Initiative »
Programs & resources for
educators, schools & students.
Computerbasedmath.org »
Join the initiative for modernizing
math education.
Step-by-Step Solutions »
Walk through homework problems one step at a time, with hints to help along the way.
Wolfram Problem Generator »
Unlimited random practice problems and answers with built-in step-by-step solutions. Practice online or make a printable study sheet.
Wolfram Language »
Knowledge-based programming for everyone.
Powered by Wolfram Mathematica © 2017 Wolfram Demonstrations Project & Contributors  |  Terms of Use  |  Privacy Policy  |  RSS Give us your feedback
Note: To run this Demonstration you need Mathematica 7+ or the free Mathematica Player 7EX
Download or upgrade to Mathematica Player 7EX
I already have Mathematica Player or Mathematica 7+