Download PDF

isosceles triangle theorem∗
Wkbj79†
2013-03-21 23:07:05
The following theorem holds in geometries in which isosceles triangle can be
defined and in which SSS, AAS, and SAS are all valid. Specifically, it holds in
Euclidean geometry and hyperbolic geometry (and therefore in neutral geometry).
Theorem 1 (Isosceles Triangle Theorem). Let 4ABC be an isosceles triangle
such that AB ∼
= AC. Let D ∈ BC. Then the following are equivalent:
1. AD is a median
2. AD is an altitude
3. AD is the angle bisector of ∠BAC
A
B
C
D
Proof. 1 ⇒ 2: Since AD is a median, BD ∼
= CD. Since we have
• AB ∼
= AC
• BD ∼
= CD
∗ hIsoscelesTriangleTheoremi created: h2013-03-21i by: hWkbj79i version: h39523i Privacy setting: h1i hTheoremi h51-00i h51M04i
† This text is available under the Creative Commons Attribution/Share-Alike License 3.0.
You can reuse this document or portions thereof only if you do so under terms that are
compatible with the CC-BY-SA license.
1
• AD ∼
= AD by the reflexive property of ∼
=
we can use SSS to conclude that 4ABD ∼
= 4ACD. By CPCTC, ∠ADB ∼
=
∠ADC. Thus, ∠ADB and ∠ADC are supplementary congruent angles. Hence,
AD and BC are perpendicular. It follows that AD is an altitude.
2 ⇒ 3: Since AD is an altitude, AD and BC are perpendicular. Thus,
∠ADB and ∠ADC are right angles and therefore congruent. Since we have
• ∠B ∼
= ∠C by the theorem on angles of an isosceles triangle
• ∠ADB ∼
= ∠ADC
• AD ∼
= AD by the reflexive property of ∼
=
we can use AAS to conclude that 4ABD ∼
= 4ACD. By CPCTC, ∠BAD ∼
=
∠CAD. It follows that AC is the angle bisector of ∠BAC.
3 ⇒ 1: Since AD is an angle bisector, ∠BAD ∼
= ∠CAD. Since we have
• AB ∼
= AC
• ∠BAD ∼
= ∠CAD
• AD ∼
= AD by the reflexive property of ∼
=
we can use SAS to conclude that 4ABD ∼
= 4ACD. By CPCTC, BD ∼
= CD.
It follows that AD is a median.
Remark: Another equivalent condition for AD is that it is the perpendicular
bisector of BC; however, this fact is usually not included in the statement of
the Isosceles Triangle Theorem.
2