Triangle:  We just studied polygons. Triangle is a polygon with three sides. So, we could define a triangle as a plane closed figure bounded by three line segments.

A triangle is a polygon with three edges and three vertices. It is one of the most basic shapes in geometry. A triangle with vertices A, B, and C is denoted by ∆ABC.

Kind of Triangles

Classification of triangles based on the length of the sides

Scalene Triangle A triangle in which all three sides are of different lengths is called Scalene Triangle.

 

In this type of triangle:

\angle A \neq \angle B \neq \angle C

 t2
Isosceles Triangle A triangle in which two sides are of the same length is called Isosceles Triangle

In this type of triangle:

\angle B = \angle C

 t3
Equilateral Triangle A triangle in which all three sides are of the same length is called Equilateral Triangle.

 

In this type of triangle:

\angle A = \angle B = \angle C

 t4

Some of the diagrams have been adopted from https://en.wikipedia.org/wiki/Triangle

Classification of Triangles based on the angles

Acute-angled Triangle A triangle in which all the three angles are more than0^{\circ} and less than 90^{\circ}  is called acute-angled triangle. t5
Right-angles Triangle A triangle in which one of the angles is 90^{\circ} is called right-angled triangle. t6
Obtuse-angled triangle A triangle in which one of the angles is more than 90^{\circ}  but less than 180^{\circ} is called obtuse-angled triangle. t7

Some of the diagrams have been adopted from https://en.wikipedia.org/wiki/Triangle

Term related to a Triangle

Median A line segment joining the vertex to the mid-point of the opposite side of a triangle is called median.

In this vertex

A  is meeting at point D (such that BD=DC ) midpoint of BC

t8
Centroid The point of intersection of three medians is called centroid.  

t9
Altitude The perpendicular drawn from the vertex to the opposite side.

Here AD   is the altitude of the triangle AD \ and\  BC is the base.

 t10
Orthocenter The intersection of the three altitudes is called the Orthocenter of the triangle.

Here A is the Orthocenter of the triangle.

 t11
Angle Bisector A line segment that bisects and interior angle of a triangle is called angle bisector.

Here AD is bisecting \angle BAC into two equal \angle BAD \ and\ \angle DAC

t12
Incentre and Incircle The point of intersection of internal angle bisectors is called the Incentre.

I is the Incentre of the triangle.

Now if you draw a circle with the center

I  in such a way that it touches all the three sides of the triangle, then that is called Incircle.

 t13
Perpendicular Bisector or Right Bisector A line bisecting any side of the triangle and perpendicular to it is called perpendicular bisector of that side of the triangle.

Here BC is being bisected by DE. BD=DC \ and\ ED\perp BC.

t14
Circumcenter  and Circumcircle The point of intersection of the perpendicular bisectors of the sides of the triangle is called Circumcenter.

Here O   is the circumcenter.

 t15
Exterior Angle and Interior Opposite Angles of a Triangle \angle ACD is the exterior angle and\angle CBA  and\angle BAC   are opposite interior angles of this exterior angle. t16

Some of the diagrams have been adopted from https://en.wikipedia.org/wiki/Triangle

\displaystyle \text{PROPERTIES OF TRIANGLES}

\displaystyle \text{A. Angle Sum Property} p1

\displaystyle \text{Theorem: The sum of the three interior angles of a triangle is } 180^\circ.

\displaystyle \text{Given: }\triangle ABC.

\displaystyle \text{To Prove: }\angle ABC+\angle BCA+\angle BAC=180^\circ.

\displaystyle \text{Construction: Through }A,\text{ draw a straight line }DE \text{ parallel to }BC.

\displaystyle \therefore DE\parallel BC.

\displaystyle AB\text{ and }AC\text{ are transversals to the parallel lines } DE\text{ and }BC.

\displaystyle \text{Proof: Since }DE\parallel BC,

\displaystyle \angle ABC=\angle BAD \qquad\text{(Alternate interior angles)}\qquad\ldots\text{(i)}

\displaystyle \text{Similarly,}

\displaystyle \angle BCA=\angle CAE \qquad\text{(Alternate interior angles)}\qquad\ldots\text{(ii)}

\displaystyle \text{Adding (i) and (ii), we get}

\displaystyle \angle ABC+\angle BCA =\angle BAD+\angle CAE.\qquad\ldots\text{(iii)}

\displaystyle \text{Adding }\angle BAC\text{ to both sides of (iii), we get}

\displaystyle \angle ABC+\angle BCA+\angle BAC =\angle BAD+\angle BAC+\angle CAE.

\displaystyle \text{Since }D,\ A,\ E\text{ are collinear,}

\displaystyle \angle BAD+\angle BAC+\angle CAE=180^\circ \qquad\text{(Angles on a straight line)}.

\displaystyle \therefore \angle ABC+\angle BCA+\angle BAC=180^\circ.

\displaystyle \text{Hence, }\angle A+\angle B+\angle C=180^\circ.

\displaystyle \\

\displaystyle \text{B. Exterior Angle Property} p2

\displaystyle \text{Theorem: If one side of a triangle is produced, the exterior angle}
\displaystyle \text{so formed is equal to the sum of the two interior opposite angles.}

\displaystyle \text{Given: }\triangle ABC,\text{ with }BC\text{ produced to }D.

\displaystyle \text{To Prove: }\angle ACD=\angle ABC+\angle BAC.

\displaystyle \text{Construction: Through }C,\text{ draw }CE\parallel AB.

\displaystyle AC\text{ and }BD\text{ are transversals to the parallel lines } AB\text{ and }CE.

\displaystyle \text{Proof: Since }AB\parallel CE,

\displaystyle \angle BAC=\angle ACE \qquad\text{(Alternate interior angles)}\qquad\ldots\text{(i)}

\displaystyle \angle ABC=\angle ECD \qquad\text{(Corresponding angles)}\qquad\ldots\text{(ii)}

\displaystyle \text{Adding (i) and (ii), we get}

\displaystyle \angle BAC+\angle ABC=\angle ACE+\angle ECD.

\displaystyle \text{But }\angle ACE+\angle ECD=\angle ACD.

\displaystyle \therefore \angle ACD=\angle ABC+\angle BAC.

\displaystyle \text{Hence, the exterior angle of a triangle is equal to the sum of}
\displaystyle \text{the two interior opposite angles.}

\displaystyle \text{RESULTS ON TRIANGLES}

\displaystyle \text{Theorem 1: If two sides of a triangle are equal, then the angles}
\displaystyle \text{opposite to the equal sides are equal.}
\displaystyle \text{In }\triangle ABC,\text{ if }AB=AC,\text{ then}
\displaystyle \angle ABC=\angle ACB.

\displaystyle \text{Theorem 2: If two angles of a triangle are equal, then the sides}
\displaystyle \text{opposite to the equal angles are equal.}
\displaystyle \text{In }\triangle ABC,\text{ if }\angle ABC=\angle ACB,\text{ then}
\displaystyle AB=AC.

\displaystyle \text{RESULTS ON INEQUALITIES}

\displaystyle \text{Theorem 1: The sum of the lengths of any two sides of a triangle} x2
\displaystyle \text{is always greater than the length of the third side.}
\displaystyle \text{Thus, in }\triangle ABC,
\displaystyle AB+BC>AC,\qquad BC+CA>AB,\qquad CA+AB>BC.

\displaystyle \text{Theorem 2: The difference between the lengths of any two sides}
\displaystyle \text{of a triangle is less than the length of the third side.}
\displaystyle \text{Thus, in }\triangle ABC,
\displaystyle |AB-AC|<BC,\qquad |BC-AB|<AC,\qquad |CA-BC|<AB.

x3

\displaystyle \text{Theorem 3: If two sides of a triangle are unequal, then the greater}
\displaystyle \text{side has the greater angle opposite to it.}
\displaystyle \text{In other words, in }\triangle ABC,
\displaystyle AB>AC\quad\Rightarrow\quad\angle ACB>\angle ABC.

\displaystyle \text{In Figure 1, }BC\text{ is the longest side. Therefore, the angle}
\displaystyle \text{opposite }BC,\text{ namely }\angle A,\text{ is the greatest angle.}
\displaystyle \therefore \angle A>\angle B\quad\text{and}\quad\angle A>\angle C.

\displaystyle \text{In Figure 2, }AB\text{ is the longest side. Therefore, the angle}
\displaystyle \text{opposite }AB,\text{ namely }\angle C,\text{ is the greatest angle.}
\displaystyle \therefore \angle C>\angle A\quad\text{and}\quad\angle C>\angle B.

\displaystyle \text{Theorem 4: If two angles of a triangle are unequal, then the side}
\displaystyle \text{opposite the greater angle is greater.}
\displaystyle \text{This is the converse of Theorem 3.}
\displaystyle \text{In other words, in }\triangle ABC,
\displaystyle \angle B>\angle C\quad\Rightarrow\quad AC>AB.

\displaystyle \text{In Figure 1, }\angle A>\angle B\text{ and }\angle A>\angle C.
\displaystyle \text{Therefore, }BC\text{, the side opposite }\angle A,\text{ is the}
\displaystyle \text{longest side of the triangle.}

\displaystyle \text{In Figure 2, }\angle C>\angle A\text{ and }\angle C>\angle B.
\displaystyle \text{Therefore, }AB\text{, the side opposite }\angle C,\text{ is the}
\displaystyle \text{longest side of the triangle.}

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