22. Construct a Triangle Given the Hypotenuse and the Length of an Angle Bisector
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.
ABC
c
AB
b
C
C
a
b
Construction
Step 1. Let have length .
BA
c
Let the point be on the perpendicular to at such that . Let be the circle with center and radius .
S
AB
B
BS=
b
C
2
σ
S
SB
Step 2. Let and be the intersections of the ray and .
P
R
AS
σ
Let the points and be on so that and . Let be the point on the extension of so that is the midpoint of .
E
F
AB
EA=PA
FA=RA
G
SB
B
GS
Step 3. Let be the intersection of the ray and the perpendicular to at .
H
AG
AB
F
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 .
M
AB
τ
M
c/2
C
τ
AB
H
Step 5. Draw the triangle .
ABC
Verification
From step 1, .
BA=c
Step 4 and Thales's theorem imply that triangle is right-angled at .
ABC
C
It remains to prove that the length of the angle bisector at is .
C
b
C
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 .
D
AB
I
J
D
CB
CA
CIDJ
CD
C
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. ■
CIDJ
CD
π/4
CB
CA
π/2
C
CD
C
CID
CJD
DI=DJ
CIDJ
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 .
ABC
c
a
b
b
C
a+b
xx-
b
C
2
=2
c
h
C
c=
h
C
b
C
2
(a+b)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 =++2ab=+(a+b), we have a quadratic equation for , namely, -x= or .Twice the area of is so . ■
CD
C
D
AB
CIDJ
b
C
2
b
C
ABC
(a+b)
b
C
2
=ab2
(a+b)
2
a
2
b
2
c
2
b
C
x=a+b
2
x
2
b
C
2
c
xx-
b
C
2
=2
c
ABC
c
h
C
c=
h
C
b
C
2
(a+b)The first equation of theorem 2 gives a construction of using the power of with respect to . In this case, by step 2, , since =FAEF=RAPA=RARA-.
a+b
A
σ
AF=AR=a+b
2
c
2
b
C
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 =HF from .
h
C
H
HFA
GBA
h
C
HF/GB=AF/AB=(a+b)/c
Since is a right triangle with hypotenuse and altitude , using theorem 2, .
ABC
c
h
C
CD=
b
C