21. Construct a Triangle with Equal Base Angles That Are Each Double the Third Angle

move B

1

2

3

4

5

6

7

8

proof

This Demonstration constructs an isosceles triangle with base angles that are double the third angle [2].

Construction

1. Draw the line segment

AB

.

2. Draw the circle

σ

with center

A

and radius

AB

.

3. Draw the perpendicular

ρ

to

AB

at

A

with

C

as one point of intersection with

σ

.

4. Let

M

be the midpoint of

AC

.

5. Let

E

be on the other side of

A

from

D

such that

DE=DB

.

6. Let

F

be on

AB

such that

AF=AE

.

7. Let

G

be on

σ

such that

GB=AF

.

8. Then

ΔAGB

meets the required conditions.

Using Euclid II.11 [3],

F

is constructed on

AB

such that

ABBF=

2

AF

=

2

BG

. The point

G

is constructed such that

BG=AF

. Since

2

BG

=BFBA

,

BG

is tangent to the circumcircle of

△AFG

(Euclid, III, 37 [4]). Hence,

∠BAG=∠FGB

. The exterior angle theorem applied to

△GAF

gives

∠BFG=∠BAG+∠FGA

. So

∠BFG=∠BGA

. Hence,

△BGF

is isosceles and

GF=BG=AF

. So,

∠FGA=∠FAG

. Then

∠BGA=2∠BAG

. Therefore,

∠BAG=

°

36

,

∠AGB=

°

72

.