# 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