Lines and Angles: 9 Math
Theorems on Lines and angles : 9th math
Theorem 6.1 (Axiom: 6.1)
If a ray stands on a line, then the sum of the adjacent angles so formed is 180o.
Prove
Given,
Let PQ is a given straight line
And given line PS stands on the straight line AB
Thus, to prove: ∠POS + ∠SOQ = 180o
Construction
A line RO `_|_` PQ is drawn
Proof
Here, from the figure we have
∠POS = ∠POR + ∠ROS - - - - - (i)
And, ∠QOS = ∠QOR – ∠ROS - - - - - - (ii)
After adding equation (i) and equation (ii), we get
∠POS + ∠QOS = ∠POR + ∠ROS + ∠QOR – ∠ROS
⇒ ∠POS + ∠QOS = ∠POR + ∠QOR
⇒ ∠POS + ∠QOS = 90o + 90o
[Because RO `_|_` PQ]
⇒ ∠POS + ∠QOS = 180o Proved
If a ray stands on a line, then the sum of the adjacent angles so formed is 180o.
Theorem: 6.2 (Axiom 6.2)
Converse of Theorem: 6.1 (Converse of Axiom 6.1)
If sum of two adjacent angles is 180o, then their non common arms of the angles form a line.
Given,
Two adjacent angles ∠QOR and ∠ROP
And, ∠QOR + ∠ROP = 180o
Then, to prove: OQ and OP form opposite angles that mean OQ and OP form a straight line.
Construction:
A line OS is drawn on PQ such that POS is a straight line.
Proof:
Since POS is a straight line,
Therefore, ∠ROS + ∠ROP = 180o - - - - - (i)
And ∠QOR + ∠ROP = 180o - - - - - (ii)
[As given in question]
Thus, from equation (i) and equation (ii), we get
∠ROS + ∠ROP = ∠QOR + ∠ROP
[Because things which are equal to the same are equal to one another]
⇒ ∠ROS + ∠ROP – ∠ROP = ∠QOR
Now, if angle ∠ROS and ∠QOR are equal, it is only possible if OS and OQ are the same line.
Therefore, it can be said that OP and OQ are opposite to each other.
Hence, If sum of two adjacent angles is 180o, then their non common arms of the angles form a line. Proved
Theorem 6.3
The sum of all angles around a Point is equal to 360o
Given, OP, OQ, OR, OS and OT are rays and these are making angles around point O.
Thus, to prove
The sum of all angles around point O = 360o
i.e. ∠POQ + ∠QOP + ∠ROS + ∠SOT + ∠TOP = 360o
Construction
Let a ray OU is drawn opposite to ray OP.
Proof
We know that, If a ray stands on a line, then the sum of the adjacent angles so formed is 180o.
Here, Ray OQ is standing on Line UP
Therefore, ∠POQ + ∠QOU = 180o
[As ∠POQ and ∠QOU are linear pair of angles.]
⇒ ∠POQ + ∠QOR + ∠ROU = 180o - - - - - (i)
[Because ∠QOP + ∠ROU = ∠QOU]
Similarly, ray OS stands on line UP
Therefore, ∠UOS + ∠SOP = 180o
⇒ ∠UOS + ∠SOT + ∠TOP = 180o - - - - - (ii)
[Because ∠SOT + ∠TOP = ∠SOP]
Now, after adding equation (i) and equation (ii), we get
∠POQ + ∠QOR + ∠ROU + UOS + ∠SOT + ∠TOP = 360o
⇒ ∠POQ + ∠QOR + ∠ROS + ∠SOT + ∠TOP = 360o
[Because, ∠ROU + ∠UOS = ∠ROS]
Therefore, ∠POQ + ∠QOP + ∠ROS + ∠SOT + ∠TOP = 360o Proved
Thus, The sum of all the angles around a point is equal to 360o Proved
Theorem 6.4
If two lines intersect each other, then the vertically opposite angles are equal.
Given, two lines intersect each other.
Let two lines PQ and RS intersect each other at point O
Then, to prove ∠POS = ∠ROS
And, ∠POR = ∠SOQ
Proof
Since, ray OP stands on line RS
Therefore, according to Axiom of linear Pair which says that If a ray stands on a line, then the sum of the adjacent angles so formed is 180o
∠POR + ∠POS = 180o - - - - - - (i)
Similarly, since say OR stands on line PQ
Therefore, ∠POR + ∠ROQ = 180o - - - - - - (ii)
Now, from equation (i) and equation (ii), we get
∠POR + ∠POS = ∠POR + ∠ROQ
Now, by transposing ∠POR to LHS
⇒ ∠POR + ∠POS – ∠POR = ∠ROQ
⇒ ∠POS = ∠ROQ
Similarly, since ray OS stands on line PQ
Thus, ∠POS + ∠SOQ = 180o - - - - - - (iii)
Now, from equation (i) and equation (iii)
∠POR + ∠POS = ∠POS + ∠SOQ
⇒ ∠POR + ∠POS – ∠POS = ∠SOQ
⇒ ∠POR = ∠SOQ
Thus, ∠POS = ∠ROQ and ∠POR = ∠SOQ Proved
Thus, If two lines intersect each other, then the vertically opposite angles are equal. Proved.
Theorem 6.5
If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal
Given, A transversal intersects two parallel line and a transversal intersects them at two points.
Let AB and CD are two parallel lines
And a transversal RS intersects these two parallel lines at points P and Q
Then to prove ∠APQ = ∠PQD
And, ∠QPB = ∠CQP
Proof
We know that according to one of the theorem of parallel lines which says if two lines intersects each other, then the vertically opposite angles are equal.
Therefore, ∠APQ = ∠RBP - - - - - (i)
[Because ∠APQ and ∠RBP are vertically opposite angles]
And according to the axioms of corresponding angles which says if a transversal intersects two parallel lines, the each pair of corresponding angles is equal.
Therefore, ∠RPB = ∠PQD - - - - - (ii)
Now, from equation (i) and equation (ii), we get
Similarly, ∠QPB = ∠APR - - - - (iii)
[Because ∠QPB and ∠APR are vertically opposite angles and hence are equal]
And, ∠CQP = ∠APR - - - - - - (iv)
[Because ∠CQP and ∠APR are corresponding angles of made by two parallel lines and a intersecting transversal, and hence are equal]
Now, from equation (iii) and equation (iv), we get
Thus,
Thus, If a transversal intersects two parallel lines, then each pair of alternate angles are equal
Theorem 6.6 (Converse of theorem 6.5)
If a transversal intersects two lines such that a pair of alternate interior angles is equal, the two lines are parallel.
Given, A transversal intersects two parallel lines.
And a pair of alternate angles is equal.
Let AB and CD are two parallel lines.
And a transversal RS intersect these parallel lines at points P and Q.
And, ∠APQ = ∠PQD
Then, to prove AB||CD.
Proof
We know that according to one of the theorem of parallel lines which says if two lines intersects each other, then the vertically opposite angles are equal.
Thus, ∠APQ = ∠RPB - - - - - (i)
[Because ∠APQ and ∠RPB are vertically opposite angles and hence are equal.]
Again as given in the question,
∠APQ = ∠PQD - - - - - (ii)
Now, from equation (i) and equation (ii), we get
This means here pair of corresponding angles is equal.
Now, according to one of the axioms of parallel lines which says that If a transversal intersects two lines such that a pair of corresponding angles is equal, the two lines are parallel to each other.
Thus, AB||CD Proved
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