3. Construct a Triangle Given the Length of an Altitude, the Inradius and the Difference of the Base Angles

h

C

1

r

0.4

δ

0.6

step 1

step 2

step 3

step 4

This Demonstration constructs a triangle

ABC

given the length

h

C

of the altitude from

C

to the base

AB

, the inradius

r

and the difference

δ

of the angles at the base.

Construction

Let

M

be a point on a line that will contain the base

AB

. Let

C

be on the perpendicular to

M

with

CM=

h

C

. Let

CN

be the straight line such that

∠MCN=δ/2

with

N

on the base.

Step 1: Let

D

be a point on

CM

such that

DM=r

. Let

S

be the intersection of

CN

and the perpendicular to

CM

at

D

; therefore, the distance from

S

to

MN

is

r

. Let

σ

be the circle with center

S

and radius

r

. The circle

σ

is tangent to

MN

.

Step 2: Draw a circle

τ

with

CS

as a diameter.

Step 3: Join

C

to both intersections of

σ

and

τ

. These two lines are tangent to

σ

; extend them to intersect

MN

at

A

and

B

.

Step 4: Draw the triangle

ABC

.

Verification

Let

∠CAB=α

,

∠ABC=β

and

∠BCA=γ

.

Theorem: In any triangle

ABC

, if

α≥β

, then the angle between the altitude from

C

and the angle bisector at

C

is

(α-β)/2

.

Proof:

∠MCN=∠NCA-∠MCA=γ/2-(π/2-α)=(π-(α+β))/2-(π/2-α)=(α-β)/2

.

By construction, the altitude from

C

has length

h

C

. The circle

σ

is the incenter of

ABC

by steps 1 and 3.

By construction,

∠MCN=δ/2

, so

δ/2=(α-β)/2

; that is,

δ=α-β

.