Lines and Angles
Parallel Lines And A Transversal
Theorem 4
If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.
Solution:
Given: Transversal EF intersects two parallel lines AB and CD at G and H respectively.
To Prove: ∠1 + ∠4 = 180° and ∠2 + ∠3 = 180°
Proof: ∠2 + ∠5 = 180° ………equation (i) (Linear pair of angles)
But ∠5 = ∠3 ……………equation (ii) (corresponding angles)
From equations (i) and (ii),
∠2 + ∠3 = 180°
Also, ∠3 + ∠4 = 180° ………equation (iii) (Linear pair)
But ∠3 = ∠1 …………..equation (iv) (Alternate interior angles)
From equations (iii) and (iv)
∠1 + ∠4 = 180° and ∠2 + ∠3 = 180° Proved
Theorem 5
If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.
Solution:
Given: A transversal EF intersects two lines AB and CD at P and Q respectively.
To Prove: AB ||CD
Proof: ∠1 + ∠2 = 180° ………..equation (i) (Given)
∠1 + ∠3 = 180° …………..equation (ii) (Linear Pair)
From equations (i) and (ii)
∠1 + ∠2 = ∠1 + ∠3
Or, ∠1 + ∠2 - ∠1 = ∠3
Or, ∠2 = ∠3
But these are alternate interior angles. We know that if a transversal intersects two lines such that the pair of alternate interior angles are equal, then the lines are parallel.
Hence, AB║CD Proved.
Theorem 6
Lines which are parallel to the same line are parallel to each other.
Solution:
Given: Three lines AB, CD and EF are such that AB║CD, CD║EF.
To Prove: AB║EF.
Construction: Let us draw a transversal GH which intersects the lines AB, CD and EF at P, Q and R respectively.
Proof: Since, AB║CD and GH is the transversal. Therefore,
∠1 = ∠2 ………….equation (i) (corresponding angles)
Similarly, CD ||EF and GH is transversal. Therefore;
∠2 = ∠3 ……………equation (ii) (corresponding angles)
From equations (i) and (ii)
∠1 = ∠3
But these are corresponding angles.
We know that if a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines ate parallel to each other.
Hence, AB║ EF Proved.
Angle Sum Property of Triangle:
Theorem 7
The sum of the angles of a triangle is 180º.
Solution:
Given: Δ ABC.
To Prove: ∠1 + ∠2 + ∠3 = 180°
Construction: Let us draw a line m though A, parallel to BC.
Proof: BC ||m and AB and AC are its transversal.
Hence, ∠1 = ∠4 …………….equation (i) (alternate interior angles)
∠2 = ∠5 ………..equation (ii) (alternate interior angles)
By adding equation (i) and (ii)
∠1 + ∠2 = ∠4 + ∠5 ………..equation (iii)
Now, by adding ∠3 to both sides of equation (iii), we get
∠1 + ∠2 + ∠3 = ∠4 + ∠5 + ∠3
Since, ∠4 + ∠5 + ∠ = 180° (Linear group of angle)
Hence, ∠1 + ∠2 + ∠3 = 180°
Hence Proved.
Theorem 8
If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.
Solution:
Given: ΔABDC in which side BC is produced to D forming exterior angle ∠ACD of ΔABC.
To Prove: ∠4 = ∠1 + ∠2
Proof: Since, ∠1 + ∠2 + ∠3 = 180°…………equation (i) (angle sum of triangle)
∠2 + ∠4 = 180° ………….equation (ii) (Linear pair)
From equations (i) and (ii)
∠1 + ∠2 + ∠3 = ∠3 + ∠4
Or, ∠1 + ∠2 + ∠3 - ∠3 = ∠4
Or, ∠1 + ∠2 = ∠4
Hence, ∠4 = ∠1 + ∠2 Proved