converse of isosceles triangle theorem

converse of isosceles triangle theorem

The following

theorem

holds in

geometries

in which

isosceles triangle

can be defined and in which SAS, ASA, and AAS are all valid. Specifically, it holds in

Euclidean geometry

and

hyperbolic geometry

(and therefore in

neutral geometry

).

Theorem 1

().

If

△

⁢

A

⁢

B

⁢

C

is a

triangle

with

D

∈

B

⁢

C

¯

such that any two of the following three statements are true:

1.

A

⁢

D

¯

is a median

2.

A

⁢

D

¯

is an

altitude

3.

A

⁢

D

¯

is the

angle bisector

of

∠

⁢

B

⁢

A

⁢

C

then

△

⁢

A

⁢

B

⁢

C

is isosceles.

Proof.

First, assume 1 and 2 are true. Since

A

⁢

D

¯

is a median,

B

⁢

D

¯

≅

C

⁢

D

¯

. Since

A

⁢

D

¯

is an altitude,

A

⁢

D

¯

and

B

⁢

C

¯

are

perpendicular

. Thus,

∠

⁢

A

⁢

D

⁢

B

and

∠

⁢

A

⁢

D

⁢

C

are

right angles

and therefore

congruent

. Since we have

•

A

⁢

D

¯

≅

A

⁢

D

¯

by the reflexive property (

http://planetmath.org/

Reflexive

) of

≅

•

∠

⁢

A

⁢

D

⁢

B

≅

∠

⁢

A

⁢

D

⁢

C

•

B

⁢

D

¯

≅

C

⁢

D

¯

we can use SAS to conclude that

△

⁢

A

⁢

B

⁢

D

≅

△

⁢

A

⁢

C

⁢

D

. By CPCTC,

A

⁢

B

¯

≅

A

⁢

C

¯

.

Next, assume 2 and 3 are true. Since

A

⁢

D

¯

is an altitude,

A

⁢

D

¯

and

B

⁢

C

¯

are perpendicular. Thus,

∠

⁢

A

⁢

D

⁢

B

and

∠

⁢

A

⁢

D

⁢

C

are right angles and therefore congruent. Since

A

⁢

D

¯

is an angle bisector,

∠

⁢

B

⁢

A

⁢

D

≅

∠

⁢

C

⁢

A

⁢

D

. Since we have

•

∠

⁢

A

⁢

D

⁢

B

≅

A

⁢

D

⁢

C

•

A

⁢

D

¯

≅

A

⁢

D

¯

by the reflexive property of

≅

•

∠

⁢

B

⁢

A

⁢

D

≅

∠

⁢

C

⁢

A

⁢

D

we can use ASA to conclude that

△

⁢

A

⁢

B

⁢

D

≅

△

⁢

A

⁢

C

⁢

D

. By CPCTC,

A

⁢

B

¯

≅

A

⁢

C

¯

.

Finally, assume 1 and 3 are true. Since

A

⁢

D

¯

is an angle bisector,

∠

⁢

B

⁢

A

⁢

D

≅

∠

⁢

C

⁢

A

⁢

D

.

Drop perpendiculars

from

D

to the rays

A

⁢

B

→

and

C

⁢

D

→

. the

intersections

as

E

and

F

, respectively. Since the length of

D

⁢

E

¯

is at most

B

⁢

D

¯

, we have that

E

∈

A

⁢

B

¯

. (Note that

E

≠

A

and

E

≠

B

are

not

assumed.) Similarly

F

∈

A

⁢

C

¯

.

Since we have

•

∠

⁢

A

⁢

E

⁢

D

≅

∠

⁢

A

⁢

F

⁢

D

•

∠

⁢

B

⁢

A

⁢

D

≅

∠

⁢

C

⁢

A

⁢

D

•

A

⁢

D

¯

≅

A

⁢

D

¯

by the reflexive property of

≅

we can use AAS to conclude that

△

⁢

A

⁢

D

⁢

E

≅

△

⁢

A

⁢

D

⁢

F

. By CPCTC,

D

⁢

E

¯

≅

D

⁢

F

¯

and

∠

⁢

A

⁢

D

⁢

E

≅

∠

⁢

A

⁢

D

⁢

F

.

Since

A

⁢

D

¯

is a median,

B

⁢

D

¯

≅

C

⁢

D

¯

. Recall that SSA holds when the angles are right angles. Since we have

•

B

⁢

D

¯

≅

C

⁢

D

¯

•

D

⁢

E

¯

≅

D

⁢

F

¯

•

∠

⁢

B

⁢

E

⁢

D

and

∠

⁢

C

⁢

F

⁢

D

are right angles

we can use SSA to conclude that

△

⁢

B

⁢

D

⁢

E

≅

△

⁢

C

⁢

D

⁢

F

. By CPCTC,

∠

⁢

B

⁢

D

⁢

E

≅

∠

⁢

C

⁢

D

⁢

F

.

Recall that

∠

⁢

A

⁢

D

⁢

E

≅

∠

⁢

A

⁢

D

⁢

F

and

∠

⁢

B

⁢

D

⁢

E

≅

∠

⁢

C

⁢

D

⁢

F

. Thus,

∠

⁢

A

⁢

D

⁢

B

≅

∠

⁢

A

⁢

D

⁢

C

. Since we have

•

A

⁢

D

¯

≅

A

⁢

D

¯

by the reflexive property of

≅

•

∠

⁢

A

⁢

D

⁢

B

≅

∠

⁢

A

⁢

D

⁢

C

•

B

⁢

D

¯

≅

C

⁢

D

¯

we can use SAS to conclude that

△

⁢

A

⁢

B

⁢

D

≅

△

⁢

A

⁢

C

⁢

D

. By CPCTC,

A

⁢

B

¯

≅

A

⁢

C

¯

.

In any case,

A

⁢

B

¯

≅

A

⁢

C

¯

. It follows that

△

⁢

A

⁢

B

⁢

C

is isosceles.
∎

Title

converse of isosceles triangle theorem

Canonical name

ConverseOfIsoscelesTriangleTheorem

Date of creation

2013-03-22 17:12:20

Last modified on

2013-03-22 17:12:20

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

7

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 51-00

Classification

msc 51M04

Related topic

IsoscelesTriangleTheorem

Related topic

AngleBisectorAsLocus

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