Download converse of isosceles triangle theorem

converse of isosceles triangle theorem∗
Wkbj79†
2013-03-21 23:07:23
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 (Converse of Isosceles Triangle Theorem). If 4ABC is a triangle
with D ∈ BC such that any two of the following three statements are true:
1. AD is a median
2. AD is an altitude
3. AD is the angle bisector of ∠BAC
then 4ABC is isosceles.
A
B
C
D
Proof. First, assume 1 and 2 are true. Since AD is a median, BD ∼
= CD. Since
AD is an altitude, AD and BC are perpendicular. Thus, ∠ADB and ∠ADC
are right angles and therefore congruent. Since we have
∗ hConverseOfIsoscelesTriangleTheoremi created: h2013-03-21i by: hWkbj79i version:
h39526i 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 ∼
=
• ∠ADB ∼
= ∠ADC
• BD ∼
= CD
we can use SAS to conclude that 4ABD ∼
= 4ACD. By CPCTC, AB ∼
= AC.
Next, assume 2 and 3 are true. Since AD is an altitude, AD and BC
are perpendicular. Thus, ∠ADB and ∠ADC are right angles and therefore
congruent. Since AD is an angle bisector, ∠BAD ∼
= ∠CAD. Since we have
• ∠ADB ∼
= ADC
• AD ∼
= AD by the reflexive property of ∼
=
• ∠BAD ∼
= ∠CAD
we can use ASA to conclude that 4ABD ∼
= 4ACD. By CPCTC, AB ∼
= AC.
Finally, assume 1 and 3 are true. Since AD is an angle bisector, ∠BAD ∼
=
−−→
−−→
∠CAD. Drop perpendiculars from D to the rays AB and CD. Label the
intersections as E and F , respectively. Since the length of DE is at most BD,
we have that E ∈ AB. (Note that E 6= A and E 6= B are not assumed.)
Similarly F ∈ AC.
A
E
B
Since we have
F
D
C
• ∠AED ∼
= ∠AF D
• ∠BAD ∼
= ∠CAD
• AD ∼
= AD by the reflexive property of ∼
=
we can use AAS to conclude that 4ADE ∼
= 4ADF . By CPCTC, DE ∼
= DF
∼
and ∠ADE = ∠ADF .
Since AD is a median, BD ∼
= CD. Recall that SSA holds when the angles
are right angles. Since we have
2
• BD ∼
= CD
• DE ∼
= DF
• ∠BED and ∠CF D are right angles
we can use SSA to conclude that 4BDE ∼
= 4CDF . By CPCTC, ∠BDE
∠CDF .
Recall that ∠ADE ∼
= ∠ADF and ∠BDE ∼
= ∠CDF . Thus, ∠ADB
∠ADC. Since we have
∼
=
∼
=
• AD ∼
= AD by the reflexive property of ∼
=
• ∠ADB ∼
= ∠ADC
• BD ∼
= CD
we can use SAS to conclude that 4ABD ∼
= 4ACD. By CPCTC, AB ∼
= AC.
∼
In any case, AB = AC. It follows that 4ABC is isosceles.
3