3. Construct a Triangle Given the Length of an Altitude, the Inradius and the Difference of the Base Angles
3. Construct a Triangle Given the Length of an Altitude, the Inradius and the Difference of the Base Angles
This Demonstration constructs a triangle given the length of the altitude from to the base , the inradius and the difference of the angles at the base.
ABC
h
C
C
AB
r
δ
Construction
Let be a point on a line that will contain the base . Let be on the perpendicular to with . Let be the straight line such that with on the base.
M
AB
C
M
CM=
h
C
CN
∠MCN=δ/2
N
Step 1: Let be a point on such that . Let be the intersection of and the perpendicular to at ; therefore, the distance from to is . Let be the circle with center and radius . The circle is tangent to .
D
CM
DM=r
S
CN
CM
D
S
MN
r
σ
S
r
σ
MN
Step 2: Draw a circle with as a diameter.
τ
CS
Step 3: Join to both intersections of and . These two lines are tangent to ; extend them to intersect at and .
C
σ
τ
σ
MN
A
B
Step 4: Draw the triangle .
ABC
Verification
Let , and .
∠CAB=α
∠ABC=β
∠BCA=γ
Theorem: In any triangle , if , then the angle between the altitude from and the angle bisector at is .
ABC
α≥β
C
C
(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 has length . The circle is the incenter of by steps 1 and 3.
C
h
C
σ
ABC
By construction, , so ; that is, .
∠MCN=δ/2
δ/2=(α-β)/2
δ=α-β