Parallel Lines

Given: RP ∥ line MS and line DK is their transversal.

(i) ∠ DHP + ∠ RHD = 180° (linear pair angle) means that linear pair is a pair of adjacent, supplementary angle. Adjacent means next to each other, and supplementary means that measures of the two angles add up to equal 180 °.

∠ DHP + ∠ RHD = 180°

85° + ∠ RHD = 180° (∠ DHP = 85° given)

∠ RHD = 180° -85°

∠ RHD = 95°

(ii) ∠ RHD ≅ ∠ PHG (vertically opposite angles formed are congruent)

So, ∠ PHG = 95°

(iii) line RP || line MS (given)

∠ DHP ≅ ∠ HGS (corresponding angles) if two parallel line are cut by a transversal, then the pairs of corresponding angle are congruent.

∠ DHP = 85° (given)

So, ∠ HGS = 85°

(iv) ∠ HGS ≅ ∠ MKG (vertically opposite angles formed are congruent)

So, ∠ MKG = 85°

Given line P ∥ line Q and line L and M are transversal.

To find: ∠ a, ∠ b, ∠ c∠ d.

Construction: extend G and E in answer diagram.

∠ a + ∠ e = 180° (linear pair angle) means that linear pair is a pair of adjacent, supplementary angle.

Adjacent means next to each other, and supplementary means that measures of the two angles add up to equal 180°.

∠ a + 110° = 180° (given)

∠ a = 180° -110°

∠ a = 70°

∠ a ≅ ∠ g (vertically opposite angles formed are congruent

∠ a = 70° (prove above)

∠ 70° ≅ ∠ g

Line P || line Q and line L transversals (given)

∠ g = ∠ b (corresponding angles)

∠ b = 70°

Line P || line Q and line M is transversal (given)

∠ c ≅ ∠ f (corresponding angles) if two parallel line are cut by a transversal, then the pairs of corresponding angle are congruent.

So, ∠ f = 115° (given)

Then, ∠ c = 115°

∠ d + ∠ f = 180° (linear pair angle)

∠ d + 115° = 180° (given)

∠ d = 180° -115°

∠ d = 65°

Given line L || line M and line P is transversal.

To find: ∠ a, ∠ b, ∠ c

Construction: extend E and D in answer diagram

∠ e ≅ ∠ d (corresponding angle theorem) if two parallel line are cut by a transversal, then the pairs of corresponding angle are congruent.

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∴ ∠ d = 45° (given above diagram)

∠ a + ∠ d = 180° (linear pair angle)

∠ a + 45° = 180°

∠ a = 180° -45°

∠ a = 135°

∠ a ≅ ∠ b (vertically opposite angles formed are congruent)

∴ ∠ b = 135°

Line N || line P and line M is transversal (given)

∴ ∠ b≅ ∠ c (corresponding angle theorem) if two parallel line are cut by a transversal, then the pairs of corresponding angle are congruent.

∴ ∠ c = 135°.

Given ∠ PQR AND ∠XYZ are parallel and also YZ and QR are parallel

TO find: ∠ PQR ≅ ∠ XYZ

Construction: extend sag XY such that Q-S-R.

PQ|| XY (given)

PQ || XS and QR is transversals (from construction)

∠ PQR ≅ ∠ XSR (corresponding angle theorem) if two parallel line are cut by a transversal, then the pairs of corresponding angle are congruent) ………………………. (1)

YZ||SK and XS is transversals (given)

∠ XYZ ≅ ∠ XSR (corresponding angle theorem) if two parallel line are cut by a transversal, then the pairs of corresponding angle are congruent.) …………………………(2)

Equation (1) and (2) right side is equal

So that, ∠ PQR≅ ∠ XYZ hence proved.

Given AB ||line CD and line PQ sis transversal .and∠ PRB 105°and ∠ BRT 105°

To find: ∠ ART, ∠ CTQ, ∠ DTQ, ∠ PRB

∠ ART + ∠ BRT = 180° ° (linear pair angle) means that linear pair is a pair of adjacent, supplementary angle.

Adjacent means next to each other, and supplementary means that measures of the two angles add up to equal 180 °.)

∠ ART + 105° = 180 (∠ BRT = 105 ° given)

∠ ART = 180° -105°

∠ ART = 75°

∠ ART ≅ ∠ PRQ (vertically opposite angles formed are congruent).

So, ∠ PRB = 75° (∵ ∠ ART 75°)

Line AB || line CD line PQ is transversal (given)

∠ BRT ≅ ∠ DTQ (corresponding angle theorem) if two parallel line are cut by a transversal, then the pairs of corresponding angle are congruent).

∠ DTQ = 105 (∠ BRT is 105°)

So that, ∠ DTQ = 105°

∠ ART ≅ ∠ CTQ (corresponding angle theorem) if two parallel line are cut by a transversal, then the pairs of corresponding angle are congruent).

So that, ∠ CTQ = 75° (because ∠ ART is 75°)

Given x = 71°, y = 108°.

To find: Are the lines m and n parallel or not?

X + y = 108° + 71° (already given)

= 179°

X + y ≠ 180°

They from a pair of interior angle which are not supplement.

∴ Line M is not parallel to Line N.

∠ a≅ ∠ b

To find: line L ∥ line M

Construction: extend C in figure.

∠ a≅ ∠ b (given)

∠ b≅ ∠ c (vertically opposite angle)

∠ a ≅ ∠ c (if whenever an element A is related to an element B and B is related to an element C then A is also related to c that is called transitivity)

But they form a pair of corresponding angle that are congruent.

∴ line L ∥ line M (hence proved).

Given ∠ a≅ ∠ b and ∠ x≅ ∠ y

To find: line L ∥ line N

∠ a≅ ∠ b (given)

But they form a pair of corresponding angles which are congruent

Line L ∥ line M …………... (1)

∠ x≅ ∠ y (given)

But they form a pair of alternative angle that are congruent

Line M ∥ line N (alternative angle) ………… (2)

Equation (1) and equation (2)

Line L ∥ line N (hence proved).

Given: ray ∠ BA ∥ ∠ DE and ∠ c = 50°, ∠ D = 100°.

To find: ∠ ABC.

Construction: Extend AB such that A-B-F-G.

BA ∥ DE (given)

AG∥ DE(construction) and DC is transversal.

∠ D ≅ ∠ GFC ((corresponding angle theorem) if two parallel line are cut by a transversal, then the pairs of corresponding angle are congruent).

∠d = 100° (given)

So that ∠ GFC = 100°

∠ GFC + ∠ BFC = 180(linear pair angle

∠ GFC = 100 (proved above)

∠ GFC + ∠ BFC = 180°

∠ 100 + ∠ BFC = 180°

∴ ∠ BFC = 100-180°

∠ BFC = 80°

In Δ BFC

∠ BFC + ∠ c + ∠ FBC = 180° (sum of angle of Δ)

∠ 80° + 50° + ∠ FBC = 180° (already given value above ∠ BFC and ∠ c)

∠ FBC = 180° -50°

∠ FBC = 130°

∠ ABC + ∠ FBC = 180°

∠ ABC + 50 ° = 180°

∠ ABC = 180-50°

∠ ABC = 130.

Given ray AE∥ ray BD. ray AF is bisector of ∠ EAB and ray BC is the bisector of ∠ Abd.

To find: line AF ∥ line BC

∠ EAB = 2x (ray AF bisector ∠ EAB)

When a line, shape, or angle inti two exactly equal parts is called bisector.

∠ ABD = 2y (ray BC bisector ∠ ABD)

Ray AE ∥ ray BD and Ab is transversal.

∠ EAD ≅ ∠ ABD (alternate angle) two angle formed when a line crosses two other lines, that lie on opposite side of the transversal line and on opposite relative sides of the other lines. If the two lines crossed are parallel, the alternate angles are equal.)

2x = 2y

X = y

∠ FAB∠ ABC

But they form a pair of alternate angle that are congruent.

∴ line AF ∥ line BC (hence proved)

ray PR ∥ SQ and PQ is transversal (given)

To find: AB ∥ CD

∠ RPQ ≅ ∠ PQS (alternate angle) two angle formed when a line crosses two other lines, that lie on opposite side of the transversal line and on opposite relative sides of the other lines. If the two lines crossed are parallel, the alternate angles are equal.)

X = y

∠ BPQ = 2x (ray PR bisect ∠ BPQ)

∠ PQC = 2y (ray SQ bisect ∠ PQC)

When a line, shape, or angle inti two exactly equal parts is called bisector.

X = y

2x = 2y (multiply 2 on both side)

∠ BPQ = ∠ PQC

But they form a pair of alternate angle that are congruent.

∴ AB ∥ CD (hence proved)

When a transversal intersects two parallel lines, then sum of the interior angles, formed on the same side of the transversal is 180°.

When two parallel lines cut by I third line, the third line is called the transversal .8 angles are formed when parallel lines M and N are cut by a transversal line T.

A transversal intersects two parallel line so; corresponding angle is equal so that corresponding angle is also 40°

∠ A + ∠ b + ∠ C = 180° (the sum of the measures of the interior angle of a triangle is 180°

∠ A + ∠ B + ∠ C = 180°

∠ 76 + ∠ 48 + ∠ C = 180°

∠ 124 + ∠ c = 180°

∠ C = 180-124

∠ C = 56°

If measure of one of the alternate interior angles is 75° then the measure of the other angle is 75° because two parallel are intersected by transversal.

The figure is attached below:

(i) ∠ RPB + ∠ BPQ = 90°

(ii) ∠ RPQ + ∠ BPA = 180°

(iii)∠ RPQ = ∠ BPA (congruent angle)

(iv) ∠ RPB = ∠ QPA (congruent angle)

The diagram is given below:

To find: transversal line will be perpendicular to other parallel line.

AB ∥ CD and EF is transversal both.

So, ∠ EOB = ∠ O O1D = 90°

So that transversal line will perpendicular to other parallel line also. (hence proved)

Given: value of ∠ x and ∠ y

To find: line l ∥ line m

∠ y = 180-50 ° (linear pair angle) means that linear pair is a pair of adjacent, supplementary angle. Adjacent means next to each other, and supplementary means that measures of the two angles add up to equal 180 °.)

∠ y = 130 °

Transversal line making same angle to both line.

So, that line L ∥ line M.

Given line AB ∥ CD ∥ Line EF and line QP is their transversal.

To find: ∠ X.

AB ∥ CD ∥ EF (linear pair angle) (linear pair angle) means that linear pair is a pair of adjacent, supplementary angle .Adjacent means next to each other ,and supplementary means that measures of the two angles add up to equal 180 °.

X = z (alternate interior angles) …………. (1)

Alternate interior angles are a pair of angles on the inner side of each of those two lines but on opposite side of the transversal.

Y: z = 3:7 (given)

Let the common ratio between y and z be a

X + y = 180° (co -interior angles on the same side of the transversal)

Z + y = 180° (using equation 1)

7a + 3a = 180°

10a = 180°

A = 18°

∴ x = 7a

X = 7× 18°

X = 126°

Given line Q ∥ line R and p is their transversal. and a = 80°

To find: values of F and G.

∠ a = 80°

∠ a = ∠ c = 80° (vertically opposite angle)

∠ c + ∠ f = 180° (angle made on the same side of parallel lines are supplementary means their sum is 180°)

∠ 80° + ∠ f = 180°

∠ f = 180 = 80

∠ f = 100 °

∠ f + ∠ g = 180° ° (linear pair angle) means that linear pair is a pair of adjacent, supplementary angle. Adjacent means next to each other, and supplementary means that measures of the two angles add up to equal 180.)

∠ 100 + ∠ g = 180

∠ g = 180-100

∠ g = 80°

Given: Line AB ∥ CF and line BC ∥ ED

To find: ∠ ABC = ∠ FDE

Construction: G and h in diagram.

AB ∥ CD and BC is transversal both

So, ∠ ABC = ∠ PCG (linear pair angle)

BC∥ ED and PF is transversal line to both

So, ∠ ECG = ∠ CDH

∠ CD = ∠ FDE (Both angle is opposite)

∴ ∠ ABC = ∠ FDE (hence proved)

Given: PS is transversal of parallel line AB and line CD.

To find: QXRY is rectangle.

∠ AQR + ∠ CRQ = 180°

(divide by 2)

∠ XQR + ∠ XRQ = 90°

[ QX and RX are bisector)

In Δ XQR

∠ XQR + ∠ XRQ + ∠ QXR = 180°

90° + ∠ QXR = 180° (∠ XQR + ∠ XRQ = 180° proved above)

∠ QXR = 180° -90°

∠ QXR = 90°

Similarly, ∠ QYR = 90°

∠ AQR + ∠ BQR = 180 (straight line)

(divide by 2)

∠ XQR + ∠ YQR = 90° (QX and QY are bisector ∠)

∠ XQY = 90°

Similarly, ∠ XRY = 90°

If any quadrilateral has all the angle 90° it is a rectangle, so that QXRY is rectangle.