# 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