Lines and Angles: 9 Math

mathematics Class Nine

Theorems on Lines and angles part-2: 9th math

Theorem 6.7

If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

Given,

Let a transversal RS intersects two parallel lines AB and CD.

class 9th math lines and angles theorem 6.7

Then to prove

∠APQ + ∠CQP = 180^o

And, ∠QPB + ∠PQD = 180^o

Proof

We know according to one of the axiom of pairs of angles which says that If a ray stands on a line, then the sum of two adjacent angles so formed is 180^o.

Here, PB stands on line RS.

Therefore, ∠RPB + ∠QPB = 180^o - - - - (i)

[Because ∠RPB and ∠QPB are linear pair of angles and hence are supplementary]

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.

Thus, ∠PQD = ∠RPB - - - - (ii)

[Because ∠PQD and ∠RPB are corresponding angles.]

Now, from equation (ii) after substituting ∠RPB = ∠PQD in equation (i), we get

∠PQD + ∠QPB = 180^o

⇒ ∠QPB + ∠PQD = 180^o

Similarly, ray AP stands on line RS

Therefore, ∠APR + ∠APQ = 180^o - - - - - (iii)

[Because ∠APR and ∠APQ are linear pair of angles and hence are supplementary]

And, ∠APR = ∠CQP - - - - - (iv)

[Because ∠APR and ∠CQP are corresponding angles and hence are equal]

Now, from equation (iv) after replacing ∠APR = ∠CQP in equation (iii), we get

∠CQP + ∠APQ = 180^o

⇒ ∠APQ + ∠CQP = 180^o

Thus, If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary. Proved

Theorem 6.8

If a transversal intersects two lines such that a pair of interior angles on the same side of transversal is supplementary, then the two lines are parallel.

Let, AB and CD are two given lines

And a transversal RS intersects these parallel lines at two points P and Q.

And again given, ∠APQ + ∠CQP = 180^o

class 9th math lines and angles theorem 6.8

Thus, to prove AB||CD

Proof

We know according to one of the axiom of pairs of angles which says that If a ray stands on a line, then the sum of two adjacent angles so formed is 180^o.

Here, ray stand on line RS

Therefore, ∠APR + ∠APQ = 180^o - - - - - (i)

[Because ∠APR and ∠APQ are linear pair of angles and hence are supplementary]

And, ∠APQ + ∠CQP = 180^o - - - - - (ii)

[As given in theorem]

Now from equation (i) and equation (ii), we get

∠APR + ∠APQ = ∠APQ + ∠CQP

⇒ ∠APR + ~~∠APQ~~ – ~~∠APQ~~ = ∠CQP

⇒ ∠APR = ∠CQP

Thus, here a transversal intersects two lines AB and CD such that a pair of corresponding ∠APR and ∠CQP are equal.

Thus, by the axiom of corresponding angles which says that If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel to each other.

Thus, AB||CD Proved

Theorem 6.9

Lines which are parallel to the same line are parallel to each other.

Given, AB, CD and EF are three lines.

And a transversal PQ intersects these lines.

And, AB||CD

And, EF||CD

class 9th math lines and angles theorem 6.9

Then to prove EF||AB

Proof

We know that, Corresponding Angle Axiom says that If a transversal intersects two parallel lines, then each pair of corresponding angles is equal.

In between, AB and CD

AB and CD are parallel lines and a transversal PQ intersects them.

Thus, ∠PRB = ∠RSD - - - - - - (i)

[Because these are corresponding angles and hence are equal]

And In between EF and CD

EF and CD are parallel and a transversal PQ intersects them.

Thus, ∠RSD = ∠STF - - - - - (ii)

[Because these are corresponding angles and hence are equal]

Now, from equation (i) and (ii), we get

∠PRB = ∠STF

Here, corresponding angles PRB and corresponding angle STF are equal.

Thus, According to Converse of Corresponding Angle Axiom which says that If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel to each other.

Thus, EF||AB Proved

Thus, Lines which are parallel to the same line are parallel to each other. Proved

Theorem 6.10

If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.

Given,

Let AB and CD are two lines.

And a transversal EF intersects these lines at points G and H.

And, Ray GM is bisector of angle EGB

And, Ray HN is the bisector of angle GHD

And, GM || HN

class 9th math lines and angles theorem 6.10

Thus, to prove AB||CD

Proof

Since, according to question GM is the bisector of ∠EGB

Therefore, ∠EGM = 1/2 ∠EGB - - - - - (i)

Since, HN is the bisector of ∠GHD

Therefore, ∠EGH = 1/2 ∠GHD - - - - - (ii)

Now, since GM||HN and EF is transversal which intersect them at point G and H respectively

Therefore, corresponding ∠EGM = Corresponding ∠GHN - - - - - (iii)

[Because according to corresponding angles axiom when a transversal intersect two parallel lines, the corresponding angles are equal]

Now after substituting the value of equation (i) and (ii) in equation (iii), we get

1/2 ∠EGB = 1/2 ∠GHD

⇒ ∠EGB = ∠GHD - - - - - (iv)

Here, AB and CD are two lines and a transversal EF intersects them. In which ∠EGB and ∠GHD are one of the pair of corresponding angles and are equal according to equation (iv)

Thus, by converse of Corresponding Angles Axiom which says that If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel to each other.

Thus, AB||CD Proved.

Thus, If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel. Proved

Angle Sum Property of a Triangle

Theorem 6.11

The sum of the angles of a triangle is 180^o

As given, Let ABC is a triangle.

Then to prove

∠A + ∠B + ∠C = 180^o

class 9th math lines and angles theorem 6.11

Construction

Let a line DE|| to AB is drawn.

class 9th math lines and angles theorem 6.11 angle sum property of triagnle

Now, AB||DE; and a transversal CB intersects them.

Thus, ∠ECB and ∠CBA are a pair of alternate interior angles and hence are equal.

[Because one of the theorem of parallel lines says that If a transversal intersects two parallel lines, then each pair of alternate angles is equal]

This means, ∠ECB = ∠CBA - - - - - (i)

Similarly, AB||DE; and a transversal AC intersects them

Thus, ∠DCA = ∠CAB - - - - - - (ii)

[Because these are pair of alternate interior angles and hence ∠DCA = ∠CAB]

Now, Since DCE is a line and forming a straight angle.

Thus, ∠DCA + ∠BCA + ∠ECB = 180^o - - - - (iii)

[Because straight angle = 180^o]

Now, after substituting ∠ECB = ∠CBA from equation (i) and ∠DCA = ∠CAB from equation (ii) in equation (iii), we get

∠CAB + ∠BCA + ∠CBA = 180^o Proved

Thus, The sum of the angles of a triangle is 180^o Proved

Theorem 6.12

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

Given, Let ABC is a triangle.

And, side AB is produced to point D.

Thus, to prove

∠4 = ∠1 + ∠2

class 9th math lines and angles theorem 6.12 exterior and interior angles of triangle

Now, we know that sum of the angles of a triangle = 180^o

Thus, ∠1 + ∠2 + ∠3 = 180^o - - - - - (i)

Again since ABD is line and ∠3 and ∠4 are linear pair of angles, and hence their sum will be equal to 180^o

Thus, ∠3 + ∠4 = 180^o - - - - - (ii)

Now, from equation (i) and equation (ii), we get

∠1 + ∠2 + ∠3 = ∠3 + ∠4

⇒ ∠1 + ∠2 + ∠3 – ∠3 = ∠4

⇒ ∠1 + ∠2 = ∠4

⇒ ∠4 = ∠1 + ∠2 Proved

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

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