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Mathematics 9.

Table of contents
About the triangles (reminder)
The usual notation of the basic data of triangles is shown in Figure 26.
  • All three interior angles of an acute triangle are acute angles.
  • One of the angles of a right-angled triangle is a right angle. (Figure 27)
  • One of the angles of an obtuse triangle is an obtuse angle.
The angles of the triangle
THEOREM: The sum of the interior angles of a triangle is \latex{ 180º }.
Figure 26
\latex{\alpha ,\beta ,\gamma }-interior angles
\latex{\alpha' ,\beta' ,\gamma' }-exterior angles
\latex{\alpha'}
\latex{\gamma}
\latex{\alpha}
\latex{\beta'}
\latex{\gamma'}
\latex{\beta}
\latex{ A }
\latex{ B }
\latex{ C }
\latex{ a }
\latex{ b }
\latex{ c }
Proof
Let us draw a straight line through \latex{ B } parallel with \latex{ AC }. (Figure 28)
\latex{\gamma =\gamma _{1}} (alternate angles), \latex{\alpha =\alpha _{1}} (corresponding angles), \latex{\alpha +\beta +\gamma =\alpha _{1}+\beta _{1}+\gamma _{1}=180°}
Figure 27
leg
leg
hypotenuse
\latex{ 90° }
⯁ ⯁ ⯁
It can be seen from Figure 28 that since \latex{\beta +\beta' =180'}, thus \latex{\beta '=\alpha +\gamma }.
With this, we have shown the following theorem:
THEOREM: Any exterior angle of a triangle is equal to the sum of the two interior angles not adjacent to it.
As a consequence of the previous theorem

\latex{\alpha'+\beta ' +\gamma '=\left(\beta +\gamma \right) +\left(\gamma +\alpha \right) +\left(\alpha +\beta \right) =2\left(\alpha +\beta +\gamma \right) =360°}, 

i.e. the sum of the exterior angles of a triangle is \latex{ 360º }.

Figure 28
\latex{\alpha_{1}}
\latex{\gamma_{1}}
\latex{\beta'}
\latex{\alpha}
\latex{\beta}
\latex{\gamma}
\latex{ A }
\latex{ B }
\latex{ C }
Example 1
The ratio of two interior angles of a triangle is \latex{ 4 : 9 }. The third interior angle of the triangle is \latex{\frac{1}{18} } of the straight angle greater than the smaller one of the other two angles. Let us calculate the interior and exterior angles of the triangle.
Solution
As a result from the hypotheses of the exercise the interior angles of the
triangle are \latex{4x, 9x, 4x + 10º}. Their sum is \latex{ 180º }, i.e.
\latex{4x + 9x + 4x + 10º = 180º}, from which \latex{x = 10º}.
When substituting it we get that the interior angles of the triangle are \latex{ 40º }, \latex{ 50º }, \latex{ 90º }, and the corresponding exterior angles are \latex{ 140º }, \latex{ 130º }, \latex{ 90º } respectively.
The triangle inequality
The sum of any two sides of a triangle is greater than the third side, i.e. 
\latex{a + b \gt c; b + c \gt a; c + a \gt b.}
If the inequalities above are fulfilled for positive real numbers \latex{ a }, \latex{ b } and \latex{ c }, then a triangle can be constructed from line segments with lengths \latex{ a }, \latex{ b } and \latex{ c }.
Figure 29
\latex{ c }
\latex{ b }
\latex{ a }
These inequalities can also be expressed as follows:
\latex{a \gt |b- c|; b \gt |a-c|; c \gt |a-b|},
i.e. any side of a triangle is greater than the (absolute) difference of the other two sides.
Example 2
Is there a triangle where the ratio of the lengths of the sides is
  1.  \latex{ 3 : 6 : 8 };
  2.  \latex{ 3 : 6 : 10 ?}
Solution
  1. According to the hypotheses the sides of the triangle – measured in a length unit – are\latex{ 3x,\, 6x,\, 8x, } where \latex{ x } is a positive number. Since the sum of the length of any two sides is greater than the length of the third side:
  2. \latex{3x + 6x \gt 8x;\; 3x + 8x \gt 6x;\; 6x + 8x \gt 3x,} there is such a triangle.
 
  1. Since \latex{3x + 6x \lt 10x}, there is no such triangle.
⯁ ⯁ ⯁
Figure 30
leg
leg
base
\latex{ AB=AC=b }
\latex{ a }
\latex{ B }
\latex{ C }
\latex{ A }
\latex{ b }
\latex{ b }
A triangle is an isosceles triangle if two of its sides are equal in length. (Figure 30)
A triangle is an equilateral triangle if all of its sides are equal in length. (Figure 31)
Figure 31
\latex{ C }
\latex{ a }
\latex{ A }
\latex{ B }
\latex{ a }
\latex{ a }