The Triangles and Its Properties

1.0Triangle

A closed figure formed by joining three non-collinear points is called a triangle. The three sides and three angles of a triangle are collectively known as elements of the triangle.

2.0Classification of triangles

According to sides

(i) Scalene triangle: A triangle with three unequal sides is called a scalene triangle. $(a\ne b\ne c)$

(ii) Isosceles triangle: A triangle with any two pair of equal sides is called an isosceles triangle. $PQ=PR$, so a $△PQR$ is an isosceles triangle.

(iii) Equilateral triangle: A triangle with all of its three sides equal is called an equilateral triangle. $\mathrm{DE}=\mathrm{EF}=\mathrm{DF}$, so a $△\mathrm{DEF}$ is an equilateral triangle.

According to angles

(i) Acute angled triangle: A triangle with all three acute angles is an acute angled triangle. $0^{\circ }<(\angle \mathrm{A},\angle \mathrm{B},\angle \mathrm{C})<90^{\circ }$ so $△\mathrm{ABC}$ is an acute angled triangle.

(ii) Obtuse angled triangle: A triangle with one obtuse angle and two acute angles is called an obtuse angled triangle. $\left(90^{\circ }<\angle \mathrm{Q}<180^{\circ }\right)$ and $0^{\circ }<(\angle \mathrm{P},\angle \mathrm{R})<90^{\circ }$, so $\Delta$ $PQR$ is an obtuse angled triangle.

(iii) Right angled triangle: A triangle with one right angle and two acute angles is called a right-angled triangle. $\angle \mathrm{E}=90^{\circ }$ and $0^{\circ }<(\angle \mathrm{D},\angle \mathrm{F})<90^{\circ }$ so $△\mathrm{DEF}$ is a rightangled triangle.

3.0Median of a triangle

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

In $△ABC,D$ is the midpoint of $BC$ and $AD$ is a median. Similarly, E and F are the mid points of $AC$ and $AB$ respectively and $BE$ and CF are the medians of the triangle. The point where the three medians of a triangle meet is called the centroid of a triangle. The centroid of the triangle divides each median in the ratio 2 : 1, i.e. $\mathrm{AG}:\mathrm{GD}=\mathrm{BG}:\mathrm{GE}=\mathrm{CG}:\mathrm{GF}=2:1$

4.0Altitude of a triangle

The perpendicular line segment drawn from any vertex of a triangle to its opposite side is called an altitude.

Here AD, BE and CF are the three altitudes of a triangle ABC . The point at which the three altitudes of a triangle meet is called the orthocentre. Here point $P$ is the orthocentre of triangle $ABC$.

Q. In $△PQR,D$ is the mid-point of $QR$. (i) PM is the (ii) PD is the (iii) Is QM = MR? (iv) Which among PQ, PM, PD and PR is the shortest?

• Explanation (i) $\mathrm{PM}$ is the altitude. (ii) PD is the median. (iii) $\mathrm{No},\mathrm{QM}\ne \mathrm{MR}$. (iv) PM , as it is the perpendicular distance.

5.0Angle bisector of a triangle

A line segment which bisects any of the interior angles of a triangle is called its angle bisector.

BD bisects $\angle \mathrm{ABC},\mathrm{So},\mathrm{BD}$ is the angle bisector $\therefore \angle \mathrm{ABD}=\angle \mathrm{CBD}$

The point at which the three angle bisectors of a triangle meet is called the incentre. Here, AF and CE are also the angle bisectors of $\angle \mathrm{BAC}$ and $\angle \mathrm{BCA}$. The point of intersection of $AF,BD$ and $CE$ is called the incentre (I) of $△ABC$.

• Angle bisector in geometry is a line, ray, or segment that divides an angle into two equal angles of the same measure.
• Angle sum property of a triangle is the special property of a triangle that is used to find the value of an unknown angle in the triangle.

6.0Angle sum property of a triangle

• Draw a triangle. Cut on three angles. Rearrange them as shown in figure. The three angles now constitute one angle. This angle is a straight angle and so it has measure $180^{\circ }$.
  
  Thus, the sum of the measures of the three angles of the triangle is equal to $180^{\circ }$.
• Take three copies of any triangle say $△ABC$ as shown in below figures (i), (ii) and (iii).
  
  Now, arrange them as shown in figure (iv)
  
  These three triangles now constitute one angle after arranging as shown in figure (iv). This angle is a straight angle and so it has measure equal to $180^{\circ }$. Thus, the sum of the measures of three angles of a triangle is $180^{\circ }$. i.e., $\angle 1+\angle 2+\angle 3=180^{\circ }$
• Take a piece of paper and cut out a triangle say $△ABC$. Make the altitude $AD$ by folding $△ABC$ such that it passes through A. Fold now the three corners such that all the three vertices A, $B$ and $C$ touch at D.
  
  You find that all the three angles form together a straight angle. This again shows that the sum of the measures of the three angles of a triangle is equal to $180^{\circ }$.
• Draw any three triangles say $△\mathrm{ABC},△\mathrm{DEF}$ and $△\mathrm{PQR}$ in your note book. Use your protractor and measure each of the angles of these triangles. Tabulate your results.

Allowing marginal errors in measurement, you will find that the last column always gives $180^{\circ }$ (or nearly $180^{\circ }$ ).

• Angle sum property of a $△ABC$
  
  $\therefore \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }$
• Parallel lines $\ell$ and $m$ have been intersected by a transversal $x$.
  

• The alternate angles are equal. $\angle 4=\angle 6;\angle 3=\angle 5;\angle 1=\angle 7\text{ and }\angle 2=\angle 8$
• The corresponding angles are equal. $\angle 1=\angle 5;\angle 2=\angle 6;\angle 3=\angle 7\text{ and }\angle 4=\angle 8$
• The sum of the interior angles on the same side of the transversal is $180^{\circ }$. $\angle 3+\angle 6=180^{\circ }=\angle 4+\angle 5$

Angle sum property

Theorem: The sum of the angles of a triangle is $180^{\circ }$. Proof: Consider a $△\mathrm{ABC}$, Through A, draw a line XY parallel to BC and mark the angles as shown in figure.

Now, $\mathrm{XY}\Vert \mathrm{BC}$ and the transversal AB cuts them. $\therefore \angle 1=\angle 4$ [alternate angles] Again, $\mathrm{XY}\Vert \mathrm{BC}$ and the transversal AC cuts them. $\therefore \angle 2=\angle 5$ Now, $\angle 1+\angle 3+\angle 2=\angle 4+\angle 3+\angle 5$ [alternate angles] But, $\angle 4+\angle 3+\angle 5=180^{\circ }$ [ $\angle 1=\angle 4$ and $\angle 2=\angle 5$ ] [ XY is a straight line] $\therefore \angle 1+\angle 3+\angle 2=180^{\circ }$ Hence, the sum of the angles of a triangle is equal to $180^{\circ }$

Q. The base angle of an isosceles triangle is $15^{\circ }$ more than its vertical angle. Find each angle of the triangle.

• Explanation Let vertical angle be $x^{\circ }$ $\therefore$ Base angles of an isosceles triangle $=(x+15{)}^{\circ }$ $\Rightarrow \mathrm{x}+\mathrm{x}+15^{\circ }+\mathrm{x}+15^{\circ }=180^{\circ }$ $\Rightarrow 3\mathrm{x}+30^{\circ }=180^{\circ }$ $\Rightarrow 3\mathrm{x}=180^{\circ }-30^{\circ }=150^{\circ }$ $\Rightarrow \mathrm{x}=\frac{150^{\circ }}{3}=50^{\circ }$ Vertical angle $=50^{\circ }$ Base angle $=x+15^{\circ }=(50+15{)}^{\circ }$ Base angle $=65^{\circ }$

Q. In $△ABC,M$ is the mid-point of $BC$, length of $AM$ is 9 unit, $N$ is a point on $AM$ such that $\mathrm{MN}=1$. What is the distance of N from the centroid of the triangle?

• Explanation $M$ is the mid-point of $BC$. $\Rightarrow AM$ is the median of $△ABC,G$ is the centroid of $△ABC$ $\Rightarrow \mathrm{G}$ divides AM in the ratio $2:1$ $\Rightarrow \mathrm{AG}$ : $\mathrm{GM}=2:1$ Given, $\mathrm{AM}=9$ Let $AG=x$, then $GM=(9-x)$ $x:(9-x)=2:1$ $\Rightarrow \frac{\mathrm{x}}{9-\mathrm{x}}=\frac{2}{1}$ $\Rightarrow \mathrm{x}=2(9-\mathrm{x})$ $\Rightarrow \mathrm{x}=18-2\mathrm{x}$ $\Rightarrow 3\mathrm{x}=18$ $\Rightarrow \mathrm{x}=6$ $\therefore \mathrm{GM}=9-6=3$ Now, GN $=$ GM $-\mathrm{NM}=3-1=2$ unit

Q. In the figure, find the sum of the angles EAB, $ABC,BCD,CDE$ and DEA.

• Solution Mark the angles as shown. Then, $\angle \mathrm{EAB}+\angle \mathrm{ABC}+\angle \mathrm{BCD}+\angle \mathrm{CDE}+\angle \mathrm{DEA}$ $=(\angle 1+\angle 4+\angle 7)+\angle 8+(\angle 9+\angle 5)+(\angle 6+\angle 2)+\angle 3$ $=(\angle 1+\angle 2+\angle 3)+(\angle 4+\angle 5+\angle 6)+(\angle 7+\angle 8+\angle 9)$ [collect together angles of the three triangles] $=$ Int. $\angle \mathrm{s}$ of $△\mathrm{EAD}+$ Int. $\angle \mathrm{s}$ of $△\mathrm{ACD}+\mathrm{Int}.\angle \mathrm{s}$ of $△\mathrm{ABC}$ $=180^{\circ }+180^{\circ }+180^{\circ }=540^{\circ }$.

Q. In the figure $\mathrm{DE}\Vert \mathrm{BC},\angle \mathrm{B}=20^{\circ }$ and $\angle \mathrm{A}=30^{\circ }$. Find the lettered angles.

• Solution In $△\mathrm{ABC},30^{\circ }+20^{\circ }+\angle \mathrm{y}=180^{\circ }$ (angle sum property of a $\Delta$ ) $\Rightarrow 50^{\circ }+\angle \mathrm{y}=180^{\circ }$ $\Rightarrow \angle \mathrm{y}=180^{\circ }-50^{\circ }=130^{\circ }$ Now, $\mathrm{DE}\Vert \mathrm{BC}$ and transversal AB cuts them at D and B respectively. $\therefore \angle \mathrm{x}=\angle \mathrm{B}=20^{\circ }$ (corresponding angles) Again, $\mathrm{DE}\Vert \mathrm{BC}$ and transversal AC cuts them at E and C respectively. So, $\angle \mathrm{z}=\angle \mathrm{y}=130^{\circ }$ (Corresponding $\angle \mathrm{s}$ ) $\left[\because \angle \mathrm{y}=130^{\circ }\right]$ Finally, $\angle \mathrm{t}+\angle \mathrm{x}=180^{\circ }$ (linear pair) or $\angle \mathrm{t}+20^{\circ }=180^{\circ }$ $\left(\because \angle \mathrm{x}=20^{\circ }\right)$ $\Rightarrow \angle \mathrm{t}=180^{\circ }-20^{\circ }=160^{\circ }$.

7.0Exterior angles of a triangle

Consider a triangle $ABC$. If one side of $△ABC$, say $BC$, is produced and $X$ is any point on the ray BC , then $\angle \mathrm{ACX}$ is called an exterior angle of the $△\mathrm{ABC}$ at C . $\angle \mathrm{ACB}$ is the interior adjacent angle of $\angle \mathrm{ACX}$ while the other two angles, i.e., $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are not the interior adjacent angles. Angles A and B are called the interior opposite angles corresponding to exterior angle ACX.

Similarly, if AC is produced and Y is a point on ray AC , then $\angle \mathrm{BCY}$ is also an exterior angle of $△\mathrm{ABC}$ at C . $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are interior opposite angles of $\angle \mathrm{BCY}$ also.

Likewise, $\angle \mathrm{BAL}$ is the exterior angle of $△\mathrm{ABC}$ at A and $\angle \mathrm{B}$ and $\angle \mathrm{C}$ are its interior opposite angles.

• Exterior angle property
  
  $\therefore \angle \mathrm{ABX}=\angle \mathrm{A}+\angle \mathrm{C}$ Here, $\angle \mathrm{A}$ and $\angle \mathrm{C}$ are interior opposite angles.

Exterior angle property of a triangle

If any side of a triangle is produced, then the exterior angle so formed is equal to the sum of the interior opposite angles.

Alternate method:

Given: Consider $△\mathrm{ABC}.\angle \mathrm{ACD}$ is an exterior angle. Proof: Through C draw $\overline{\mathrm{CE}}$, parallel to $\overline{\mathrm{BA}}$.

Q. One side of a triangle is produced, and the exterior angle so formed is $120^{\circ }$. If the interior opposite angles be in the ratio 3:5, find the measure of each angle of the triangle.

• Explanation Let the given interior opposite angles be ( 3 x${)}^{\circ }$ and $(5\mathrm{x}{)}^{\circ }$. We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles. $\therefore 3\mathrm{x}+5\mathrm{x}=120^{\circ }$ $\Rightarrow 8\mathrm{x}=120^{\circ }$ $\Rightarrow \mathrm{x}=15^{\circ }$ $\therefore \angle \mathrm{A}=(3\times 15{)}^{\circ }=45^{\circ },\angle \mathrm{B}=(5\times 15{)}^{\circ }=75^{\circ }$ But, $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }$ (Angle sum property of a triangle) $\therefore 45^{\circ }+75^{\circ }+\angle \mathrm{C}=180^{\circ }$ $\Rightarrow 120^{\circ }+\angle \mathrm{C}=180^{\circ }$ $\Rightarrow \angle \mathrm{C}=180^{\circ }-120^{\circ }=60^{\circ }$. $\therefore \angle \mathrm{A}=45^{\circ },\angle \mathrm{B}=75^{\circ }$ and $\angle \mathrm{C}=60^{\circ }$.

Q. If all sides of a triangle are produced in order to make exterior angles, prove that the sum of the exterior angles so formed is $360^{\circ }$.

• Explanation Let the sides $BC,CA$ and $AB$ of $△ABC$ be produced to $D,E$ and $F$ respectively, as shown in figure. We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles. $\therefore \angle 1=\angle \mathrm{A}+\angle \mathrm{B}$ $\angle 2=\angle \mathrm{B}+\angle \mathrm{C}$ $\angle 3=\angle \mathrm{C}+\angle \mathrm{A}$ On adding the corresponding sides of (i), (ii) and (iii), we get $\angle 1+\angle 2+\angle 3=2(\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C})$ $=2\times 180^{\circ }$ [ sum of the angles of a triangle is $180^{\circ }$ ] $=360^{\circ }$ $\therefore \angle 1+\angle 2+\angle 3=360^{\circ }$ Hence, the sum of the exterior angles is $360^{\circ }$.

• The exterior angle and its adjacent angle follow the linear property i.e., the sum of the exterior angle and its adjacent angle is 180 degrees.

8.0Pythagoras theorem for right-angled triangles

A right-angled triangle has one right angle and two acute angles. The side opposite to the right angle is called the hypotenuse. The other two sides are called its legs.

In a right-angled triangle, the sides which contain the right angle are usually referred to as the base and perpendicular.

Pythagoras theorem states that the square of the length of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the lengths of the other two sides. In figure right-angled $△ABC$ has

Q. In an equilateral triangle, the median, angle bisector, altitude and perpendicular bisector of sides are all represented by the same lines. The lengths of the sides of a right-angled triangle are 5 m and 12 m . Find the length of the hypotenuse.

• Solution Square of the hypotenuse = Sum of the squares of the other two legs $=5^2+12^2=25+144=169$ $\therefore$ (Hypotenuse ${}^2=169=13\times 13=13^2$ So, Hypotenuse $=13\mathrm{m}$

Q. The hypotenuse of a right-angled triangle is 15 cm. One of its legs is 9 cm. What is the length of the other side?

• Solution Let the length of the unknown side be x . Then, $x^2+9^2=15^2$ ${\mathrm{x}}^2+81=225$ $x^2=225-81$ $x^2=144=12\times 12=12^2$ $\Rightarrow \mathrm{x}=12\mathrm{cm}$

Q. A flight of steps of length $5\mathbf{m}$ lead up to the door of a house on the first floor of a building. The horizontal distance of the first step at the base of the flight of stairs is 3 m as shown in the figure. At what height is the door from the ground?

• Solution Let the height of the door be at height x metres from the ground. So, the steps, the height of door, and the horizontal distance form a right-angled triangle as shown along side: According to Pythagoras theorem, ${\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2$ $5^2=x^2+3^2$ $x^2=5^2-3^2$ = $25-9$ $x^2=16=4^2$
  
  $\Rightarrow \mathrm{x}=4\mathrm{m}$
• If the angles of a right triangle are $30^{\circ },60^{\circ }$ and $90^{\circ }$, the hypotenuse is equal to twice the side opposite to the $30^{\circ }$ angle, i.e. $AC=2BC$.
  

9.0Triangle inequality property

The sum of any two sides of a triangle is greater than the third side. If $a,b$ and $c$ are the lengths of the triangle $PQR$ then

Q. 0 is a point in the exterior of $△ABC$. What symbol ' $>$ ', '<' or ' $=$ ' will you use to complete the statement: $0\mathrm{A}+\mathrm{OB}..\mathrm{AB}$ ? Write two other similar statements and show that: $\mathrm{OA}+\mathrm{OB}+\mathrm{OC}>\frac{1}{2}(\mathrm{AB}+\mathrm{BC}+\mathrm{CA})$.

• Explanation In $△OBA,OA+OB>AB$ (Sum of any two sides of a $\Delta >$ third side) Similarly, in $△OBC,OB+OC>BC$ In $△\mathrm{OAC},\mathrm{OA}+\mathrm{OC}>\mathrm{AC}$ Adding inequalities (i), (ii) and (iii) we have, $(OA+OB)+(OB+OC)+(OA+OC)>AB+BC+AC$. i.e., $2(\mathrm{OA}+\mathrm{OB}+\mathrm{OC})>\mathrm{AB}+\mathrm{BC}+\mathrm{CA}$ or $OA+OB+OC>\frac{1}{2}(AB+BC+CA)$.