24a. Construct a Triangle Given the Length of the Altitude to the Base, the Difference of Base Angles and the Sum of the Lengths of the Other Sides

δ

0.8

s

1.4

length of altitude from C to AB

h

C

0.5

move A'

0.5

show solution

This Demonstration shows a marked-ruler (or verging) construction of a triangle

ABC

given the length

h

C

of the altitude to the base, the difference

δ

of angles at the base and the sum

s

of the lengths of the other two sides.

Construction

From a point

D

draw two rays

ρ

1

and

ρ

2

at an angle

δ

.

Let

H

be on

ρ

1

such that

HD=h

. Let

C

be the intersection of the angle bisector of the angle

ρ

1

ρ

2

and the perpendicular to

ρ

1

at

H

. Thus

∠CDH=δ/2

and

DHC

is a right triangle with hypotenuse

DC

.

Draw the line segment

A'B'

of length

s

with

C

between

A'

and

B'

and

A'

on

ρ

1

.

Move

A'

along

ρ

1

until

B'

is on

ρ

2

; call this point

F

. Then set

A=A'

.

Let

B

be the reflection of

A

in

DC

.

Then

ABC

satisfies the stated conditions.

Proof

In the triangle

DAF

,

∠FDA=δ

,

∠AFD=β

, but the exterior angle

∠FAB=α

, so

δ=α-β

.

Triangles

DCF

and

DCB

are congruent, so

CB=CF

, and

BC+CA=a+b

.