17a. Construct a Triangle Given the Length of an Angle Bisector, the Angle at Another Vertex and Distance from the Third Vertex to the Angle Bisector
17a. Construct a Triangle Given the Length of an Angle Bisector, the Angle at Another Vertex and Distance from the Third Vertex to the Angle Bisector
This Demonstration shows a construction of a triangle given the angle at , the length of the angle bisector at and the length of the perpendicular from to the bisector.
ABC
γ
C
χ
A
A
p
B
Construction
Step 1: Draw a line segment of length ; will turn out to be on . Construct a point such that the triangle is isosceles (with ) such that the angle at is . Construct a circle with center and radius .
AD
χ
A
D
BC
S
ADS
AS=DS
S
2γ
S
AS
Step 2: Draw a line segment of length perpendicular to at . Let be the midpoint of . Let be the intersection of ray and the circle. Let and be the intersections of and with a line perpendicular to at , respectively.
DG
p
AD
D
E
AD
C
EG
I
J
AC
DC
DG
G
Step 3: Let be the point symmetric to with respect to , and let be the midpoint of , which is also on .
B
I
AD
K
IB
AD
Step 4: Then is a solution of the problem.
ABC
Verification
The angle at is since it is an inscribed angle over the chord , and the central angle is .
C
γ
AD
∠ASD
2γ
Triangles and are congruent, so , and are collinear and . So is the bisector of the angle at of length . Finally, .
BDK
DJG
B
D
C
IK=KB=p
AD
A
χ
A
BK=DG=p