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WOLFRAM|DEMONSTRATIONS PROJECT

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

1.4

0.5

0.8

step 1

step 2

step 3

step 4

This Demonstration shows a construction of a triangle

ABC

given the angle

, the length

of the angle bisector at

and the length

of the perpendicular from

to the bisector.

Construction

Step 1: Draw a line segment

of length

;

will turn out to be on

. Construct a point

such that the triangle

ADS

is isosceles (with

AS=DS

) such that the angle at

2γ

. Construct a circle with center

and radius

AS

Step 2: Draw a line segment

of length

perpendicular to

. 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

, respectively.

Step 3: Let

be the point symmetric to

with respect to

, and let

be the midpoint of

, which is also on

Step 4: Then

ABC

is a solution of the problem.

Verification

The angle at

since it is an inscribed angle over the chord

, and the central angle

∠ASD

2γ

Triangles

BDK

and

DJG

are congruent, so

and

are collinear and

IK=KB=p

. So

is the bisector of the angle at

of length

. Finally,

BK=DG=p

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