Angles and sides inequality theorems for triangles
Theorem 1 (
unequal sides theorem)
If in a triangle two sides are unequal, then the angle opposite to the longer side is greater than the angle opposite to the shorter side.
Proof
Let ABC be a triangle in which the side AC is longer than the side BC
(Figure 1). We need to prove that the angle ABC is greater than
the angle BAC.
On the ray CA representing the longer side, mark the segment CD
congruent to the smaller side BC (Figure 2). Since the side BC is
shorter than AC, the point D is located between points A and C.
Draw the segment BD joining points B and D. This segment is located
inside the angle ABC, therefore the angle ABC is greater than the angle DBC.
From the other side, the angle DBC is congruent to the angle BDC as
they are the angles at the base of the isosceles triangle DBC.
Note that the angle BDC is the exterior angle of the triangle ADB, and
therefore it is equal to the sum of the angles DAB and ABD (see the lesson
Sum of the interior angles of a triangle under the current topic Triangles
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Figure 1. To the Theorem 1
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Figure 2. To the proof of the Theorem 1
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in the section
Geometry in this site). This means that the angle
BDC is greater than the angle
CAB. Thus we have the chain of inequalities for the angles:
the angle
ABC is greater than the angle
DBC; the angle
DBC is equal to the angle
BDC, and the angle
BDC is greater than the angle
BAC.
This implies that the angle
ABC is greater than the angle
BAC.
The proof is completed.
Theorem 2 (
unequal angles theorem)
If in a triangle two angles are not congruent, then the side opposite to the greater angle is longer than the side opposite to the smaller angle.
Proof
Let ABC be a triangle in which the angle ABC is greater than the angle BAC
(Figure 3). We need to prove that the side AC is longer than the side BC.
Let us suppose that the opposite is true: the side AC is not longer than
the side BC. Then two cases can occur: either AC is of the same length
as BC or AC is shorter than BC.
In the first case the triangle ABC would has been isosceles and the angles
ABC and BAC would have been congruent (see the lesson Isosceles triangles
under the current topic Triangles in the section Geometry in this site).
This contradicts to the given condition.
In the second case the side BC would has been longer than AC and,
consequently, the angle BAC would has been greater than the angle ABC
in accordance to the Theorem 1 proved in this lesson above.
This contradicts to the given condition again.
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Figure 3. To the Theorem 2
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Thus the only remaining possibility is that the side
AC is longer than the side
BC. This is exactly what to be demonstrated.
Theorem 3 (
SAS inequality theorem)
If two sides of one triangle are congruent to the corresponding two sides of another triangle, and the included angle
of the first triangle is greater than the included angle of the second triangle, then the third side of the first triangle
is longer than the third side of the second triangle.
Proof
We are given two triangles ABC and DEF (Figure 4).
In these triangles the sides AB and DE are congruent; the sides AC
and DF are congruent too. The angle BAC is greater than the angle EDF.
We need to prove that the side BC is longer than EF.
Place the triangle DEF in a way that over-pose the sides AB and DE
(Figure 4). The segments AB and DE will coincide, because they are
congruent. The side DF will fall within the angle BAC as this angle is
greater than the angle EDF.
Draw the bisector AG of the angle CAF from its vertex A till the
intersection with the side BC at the point G. Join the points G and
F by a straight segment GF (Figure 4, red lines).
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Figure 4. To the Theorem 3
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The triangles
AGC and
AGF are congruent according to the
postulate 1 (SAS) of the lesson
Congruence tests for triangles, which is under the current topic
Triangles in the section
Geometry in this site. Indeed, these triangles have the common side
AG, the congruent sides
AC and
AF and the congruent included angles
CAF and
FAG.
Therefore, their corresponding sides
CG and
GF are of equal length:
CG =
GF.
The
triangle inequality for the triangle
BGF gives for the triangle sides length
EF <
BG +
GF.
From the other side,
BC =
BG +
CG =
BG +
GF.
It implies that
BC >
EF. This is exactly what to be demonstrated.
The proof is completed.
Theorem 4 (
SSS inequality theorem)
If two sides of one triangle are congruent to the corresponding two sides of another triangle, but the third side of the first triangle
is longer than the third side of the second triangle, then the angle included between two sides of the first triangle is greater than
the angle included between two sides of the second triangle.
Proof
We are given two triangles ABC and DEF (Figure 5).
In these triangles the sides AB and DE are congruent; the sides AC
and DF are congruent too. The side BC is longer than the side EF.
We need to prove that the angle BAC is greater than the angle EDF.
The proof is similar to that of the Theorem 2 above and uses the
same proof by contradiction method.
Let us suppose that the opposite is true: the angle BAC is not greater
than the angle EDF. Then two cases can occur: either the angle BAC
is congruent to the angle EDF or the angle BAC is smaller than EDF.
In the first case the triangle ABC would has been isosceles and the sides
BC and EF would have been congruent (see the lesson Isosceles triangles
under the current topic Triangles in the section Geometry in this site).
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Figure 5. To the Theorem 4
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This contradicts to the given condition.
In the second case the angle
BAC would has been greater than the angle
EDF and, consequently, the side
BC would has been longer than the side
EF
in accordance to the
Theorem 3 proved in this lesson above.
This contradicts to the given condition again.
Thus the only remaining possibility is that the angle
BAC is greater than the angle
EDF. This is exactly what to be demonstrated.
