27b. Construct a Triangle Given Its Perimeter, an Angle and the Length of the Altitude to the Side Opposite the Angle

p

2

γ

1

h

0.5

steps

1

2

3

This Demonstration constructs a triangle

ABC

given the perimeter

p

, the angle

γ

at

C

and the length

h

C

of the altitude from

C

to the base

AB

.

This construction uses a previous method to construct a triangle

CDE

given the length of the base

ED

, the opposite angle

C

and the length

h

C

of the altitude to the base.

Construction

1. Draw a triangle

CDE

given the length

p

of the base

ED

, the angle

γ/2+π/2

opposite the base and the length

h

C

of the altitude from

C

.

2. Let

A

be the intersection of

DE

and the perpendicular bisector of

EC

. Similarly, let

B

be the intersection of

DE

and the perpendicular bisector of

CD

.

3. Then the triangle

ABC

satisfies the conditions.

Verification

By step 1,

∠ECD=γ/2+π/2

. Also, since

EAC

is isosceles,

∠ECA=α/2

. Similarly,

∠BCD=β/2

. So

∠ACB=∠ECD-∠ECA-∠BCD=γ/2+π/2-α/2-β/2

.

Since

α+β+∠ACB=π

,

1/2∠ACB=π/2-(α/2+β/2)

.

Subtracting,

∠ACB-1/2∠ACB=[γ/2+π/2-α/2-β/2]-[π/2-(α/2+β/2)]

.

Simplifying,

∠ACB=γ

.

By step 1 and isosceles triangles,

ED=p

is the perimeter of

ABC

.