Angle Sum Property and Axioms - Lines and Angles, Class 9, Mathematics | Extra Documents & Tests for Class 9 PDF Download

AXIOM-1 : If a ray stands on a line, then the sum of two adjacent angles so formed is 180�.
This gives us another definition of linear pair of angles – when the sum of two adjacent angles is 180�, then they are called as linear pair of angles. In figure, a ray PQ standing on a line ∠forms a pair of adjacent angles ∠ BPQ and ∠ QPA.
The sum of the measure of the two angles is a straight angle which is equal to 180�.
The ray PB and PA are opposite rays. Thus, the two angles form a linear pair of angles. The above axiom can be stated in the reverse way as below:

AXIOM-2 : If the sum of two adjacent angles is 180�, then the non-common arms of the angles form a line. In figure, angles ∠BAC and ∠CAD are adjacent angles such that ∠BAC + ∠CAD = 180�

Then, we have
∠BAD = 180� (∵ ∠BAD = ∠BAC + ∠CAD)

 Thus, ∠BAD is a straight angle. Therefore, AB and AD are in one line. The axiom 1 and 2 are called linear pair axioms. The two axioms are reverse of each other.

(x) Vertically Opposite Angles :- If two lines intersect each other then, the pairs of opposite angles formed are called vertically opposite angles. Two lines AB and CD intersect at point O.

Then, two pairs of vertically opposite angles are formed.
One pair is ∠AOD & ∠BOC. The other pair is ∠AOC & ∠BOD.

THEOREM-1 :- If two lines intersect each other, then the vertically opposite angles are equal.
Given : Two lines AB and CD intersect at a point O.
Two pairs of vertically opposite angles are :
(i) ∠AOC and ∠BOD
(ii) ∠AOD and ∠BOC

To Prove : (i) ∠AOC =∠BOD 
(ii) ∠AOD = ∠BOC i.e. two pairs of vertically opposite equal angles.

Proof :

Ex. In figure, ∠AOC = 50^o and ∠BOC = 20^o , find ∠AOB

Sol. 

Ex. In figure,  are complementary angles. Find the value of x and hence find 

Sol. In figure, We have, ∠AOC + ∠BOC = 90�

⇒ (x+ 10)� + (2x + 5)� = 90�
⇒ 3x + 15� = 90�
⇒ 3x = 90� – 15� = 75�
⇒ x = 25�
Now, ∠AOC = (25 + 10)� = 35� and ∠BOC = {2 � 25 + 5}� = 55�

Ex. In figure, ray AD stands on the line CB, ∠BAD = (2x + 10)� and ∠CAD = (5x + 30)�, find the value of x and also write ∠BAD and ∠CAD.
 Sol. ∠BAD + ∠CAD = 180� (Linear pair of angles)

⇒ (2x + 10)� + (5x + 30)� = 180�

⇒ 7x + 40� = 180�

⇒ 7x = 180� – 40�

⇒ 7x = 140�

⇒ x = 140/7

⇒ x = 20

Now, ∠BAD = (2 � 20 + 10)� = 50�

and ∠CAD = (5 � 20 + 30)� = 130�

Ex. In figure, ray OC stands on the line AB and ∠BOC = 125�. Find reflex ∠AOC.

Sol. In figure, ∠AOC and ∠BOC are linear pair angles.

⇒ ∠AOC + ∠BOC = 180�

⇒ ∠AOC + 125� = 180�

⇒ ∠AOC = 180� – 125� = 55� (∵ ∠BOC = 125�)

Now, Reflex ∠AOC = 360� – ∠AOC = 360� – 55� = 305�

CORRESPONDING ANGLES AXIOM:- If a transversal intersects two parallel lines, then each pair of corresponding angles are equal.
 Conversely, if a transversal intersects two lines, making a pair of equal corresponding angles, then the lines are parallel.
 By using the above axiom, we can now deduce the other facts about parallel lines and their transversal.

THEOREM-2 :- If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

Given : AB and CD are two parallel lines and a transversal EF intersects them at point G and H respectively. Thus, the alternate 1

interior angles are ∠2 and∠1, and ∠3 and ∠4.

To Prove : ∠1 = ∠2 and ∠3 = ∠4

Proof :

Hence, proved.

THEOREM-3 (converse of theorem of 2) :- If a transversal intersects two lines in such a way that a pair of alternate interior angles are equal, then the two lines are parallel.

Given : A transversal 't' intersects two lines AB and CD at P and Q respectively such that ∠1 and ∠2 are a pair of alternate interior angles and ∠1 = ∠2.

To Prove : AB║CD

Proof :

Hence, proved.

THEOREM-4 :- If a transversal intersects two parallel lines, then each pair of consecutive interior angles are supplementary.
Given : AB and CD are two parallel lines. Transversal "t" intersects AB at P and CD at Q, making two pairs of 
consecutive interior angles, ∠1, ∠2 and ∠3, ∠4.

To Prove : 

(i) ∠1 + ∠2 = 180� and

(ii) ∠3 + ∠4 = 180�

Proof :

Hence, proved.

THEOREM-5 (converse of theorem 4) :- If a transversal intersects two lines in such a way that a pair of consecutive interior angles are supplementary, then the two lines are parallel.

Given : A transversal t intersect two lines AB and CD at P and Q respectively such that ∠1 and ∠2 are a pair of
consecutive interior angles, and ∠1 + ∠2 = 180�.

To Prove : AB║CD

Proof :

Hence, proved.

THEOREM-6 :- If two lines are parallel to the same line, they will be parallel to each other.

Given : Line m║ line  and line n ║ line .

To Prove : Line m║ line n.

Construction : Draw a line t transversal for the lines , m and n. 

Proof :

	STATEMENT	REASON
1.	∵  m║ ∴ ∠1 = ∠2	Corresponding angles
2.	∵ n║ ∴ ∠1 = ∠3	Corresponding angles
3.	∠1 =∠2 and ∠1=∠3 ⇒ ∠2=∠3	From (1) and (2)
4.	⇒ m║n	By converse of corresponding angles axiom

Hence, proved.

ANGLE SUM PROPERTY OF A TRIANGLE

In previous classes, we have learnt that the sum of the three angles of a triangle is 180�. In this section, we shall prove this fact as a theorem.
 

THEOREM-7 :- The sum of all the angles of a triangle is 180�.
Given : In a Δ PQR, P

∠1, ∠2 and ∠3 are the angles of Δ PQR.
To Prove : ∠1 + ∠2 + ∠3 = 180�
Construction :
Draw a line XPY parallel to QR through the opposite vertex P.

Proof :

Hence, proved.

EXTERIOR ANGLE OF A TRIANGLE
Consider the ΔPQR. If the side QR is produced to point S, then ∠PRS is called

an exterior angle of ΔPQR. 
The ∠PQR and ∠QPR are called the two interior opposite angles of the exterior ∠PRS.

THEOREM-8 :- If a side of a triangle is produced, then the exterior P

angle so formed is equal to the sum of the two interior opposite angles.
Given : In a Δ PQR, ∠1, ∠2 and ∠3 are the angles of ΔPQR, side QR is produced to S and exterior angle ∠PRS = ∠4.
To Prove : ∠4 = ∠1 + ∠2
Proof :

Hence, proved.