Axioms, Postulates And Theorems

The undefined objects in Euclid’s geometry are point, line and plane.

difference between an axiom and a postulate:

(1) Axiom is generally true for any field of science, while postulates can be specific on a particular field.

(2) It is impossible to prove from other axioms, while postulates are provable to axioms.

(a) let the number of apples in basket A = 10

And number of bananas in basket B = 10

Add 5 number of mangoes in each basket we get equal number of fruits in both of these baskets that is equal to 15.

(b) let we have a baskets of 10 apples.

If we remove 4 apples from the basket which were earlier part of that basket. The apples removed are less in number than the total apples earlier present.

Hence, we can say the whole is greater than the part.

There are certain elementary statement, which are self evident and which are accepted without any question in such statments we need axioms. Also axioms are needed because these statements are also applicable to other areas of mathematics and science.

Closure Property of natural numbers: Let there be two natural numbers be x and y

Then according to closure property of natural numbers under addition

If a is a natural number and b is a natural number then a + b is also a natural number.

Now this is something that can be proved by giving examples.

A natural number is set of whole numbers excluding zero, so all the positive integers are Natural numbers.

And when positive natural number is added to another positive natural number we will have a positive integer only.

Let a = 2 and b = 99

Then, a + b = 101 which is also a natural number

You can take any two natural numbers and repeat the above process, addition of those numbers will always be a natural number.

Three straight lines which do not pass through a fixed point can be drawn as:

Where AB, CD and EF are three straight lines. And G is a fix point. All the three lines do not pass through the given fixed point.

Where A is a point and AB, AC and AD are rays emanating from point A.

Angle between any two adjacent rays is acute i.e. < 90° .

∠BAC < 90°

And ∠DAC < 90°

Here, ∠1 and ∠2 are not adjacent angles. But they are supplementary angles because they two interior angles between two parallel lines.

i.e. ∠1 + ∠2 = 180°

It is an equilateral triangle where all the three points A, B and C are equidistant from each other.

in this figure, ∠1 and ∠2 are making a linear pair.

i.e. ∠1 + ∠2 = 180°

hence, ∠1 and ∠2 are straight angles.

As we see, in the given figure the lines are perpendicular to each other.

Hence, ∠YRP = ∠XOB = 90°

So, the given angles, ∠YRP and ∠XOB are right angles.

Here, in the given figure,

∠BOA = ∠1 is acute angle.

And the angle left after ∠BOA is known as reflex angle.

Hence, ∠2 is reflex angle.

Given: ∠AOC = 2x and ∠BOC = x

We know by proposition 1,

∠AOC + ∠BOC = 180°

x + 2x = 180°

3x = 180°

x = 60°

Given: ∠AOC = 90°, ∠COD = 4x and ∠BOD = x

We know by proposition 1,

∠AOC + ∠COD + ∠BOD = 180°

90° + 4x + x = 180°

5x = 180° - 90°

5x = 90°

x = 18°

Given: ∠COD = 90°, ∠COB = x and ∠BOD = x

As we know, sum of all angles around a point is 360°.

∠COD + ∠COB + ∠BOD = 360°

90° + x + x = 360°

2x = 360° - 90°

2x = 270°

x = 135°

Given: ∠AOC = x-y and ∠BOC = x + y

We know by proposition 1,

∠AOC + ∠BOC = 180°

x - y + x + y = 180°

2x = 180°

x = 90°

Given: ∠COF = 3x, ∠AOE = x and ∠BOD = x + 30

We know by proposition 4,

∠AOE = ∠BOF = x (vertically opposite angles)

We know by proposition 1,

∠COF + ∠BOF + ∠BOD = 180°

3x + x + x + 30° = 180°

5x = 180° - 30°

5x = 150°

x = 30°

Given: ∠BLC = 65° and ∠BLO = y

∠EOD = y and ∠FOD = x

We know by proposition 1,

∠BLC + ∠BLO = 180°

65° + y = 180°

y = 180°-65°

y = 115°

We know by proposition 1,

∠EOD + ∠FOD = 180°

y + x = 180°

115° + x = 180°

x = 180° - 115°

x = 65°

As we know, sum of all angles around a point is 360°. ∠BOC + ∠AOB + ∠AOD + ∠COD = 360°

120° + 50° + ∠AOD + 130° = 360°

∠AOD + 300° = 360°

∠AOD = 360° - 300°

∠AOD = 60°

In this,

∠AOD + ∠BOC = 60° + 120° = 180°

And ∠AOB + ∠COD = 50° + 130° = 180°

These are the pair of supplementary angles. because sum of angles is 180° hence these are supplementary rays.

As we know, in case of supplementary angles the sum of angles must be 180°

Let two adjacent angles are x and y.

For these angles to be supplement,

x + y = 180° …(1)

let x = 120° (obtuse angle)

put it in (1)

120° + y = 180°

⇒ y = 180° - 120°

y = 60° (i.e. acute angle)

hence, to have a pair of supplementary angle if one angle is obtuse then other must be acute angle.

Given: ∠DMP = 135°

We know by proposition 4,

∠LMC = ∠DMP = 135° (vertically opposite angles (V.O.A))

We know by proposition 1,

∠DMP + ∠DML = 180°

135° + ∠DML = 180°

∠DML = 180° - 135°

∠DML = 45°

∠LMC = ∠DML = 45° (vertically opposite angles (V.O.A))

∠BLQ = ∠DML = 45° (Corresponding angles)

∠ALM = ∠BLQ = 45° (vertically opposite angles (V.O.A))

We know by proposition 1,

∠ALM + ∠ALQ = 180°

45° + ∠ALQ = 180°

∠ALQ = 180° - 45°

∠ALQ = 135°

∠BLM = ∠ALQ = 135° (vertically opposite angles (V.O.A))

We know by proposition 1,

∠DAB + ∠BAC = 180°

130° + ∠BAC = 180°

∠BAC = 180° - 130°

∠BAC = 50°

And again We know by proposition 1,

∠BCE + ∠BCA = 180°

90° + ∠BCA = 180°

∠BCA = 180° - 90°

∠BCA = 90°

In triangle ABC,

∠ACB + ∠BAC + ∠CBA = 180°

90° + 50° + ∠CBA = 180°

∠CBA = 180° - 140°

∠CBA = 40°

∠BFG = x = ∠CBA = 40° (alteranate angles)

Let GH intersect AB and CD at L and M respectively. Since GH is perpendicular to AB. We have ∠GLA = 90°. GH is also perpendicular to CD. Thus ∠GMC = 90° . we get ∠GLA = ∠GMC = 90°. Thus the lines are parallel. And angles are corresponding angle. Since CD is also parallel with EF. Hence ∠GNE = 90°. Hence it is perpendicular to all other lines in the same way.

To Prove: ∠ 2 + ∠ 4 = 90°

The situation is shown in the diagram above

AB is parallel to CD and PQ is a traversal

From the diagram we can see that (∠1 + ∠ 2) is an interior angle and similarly ∠3 + ∠ 4 is another interior angle made on the same side

Now ∠1 = ∠ 2 (As they are angular bisector of ∠ AOQ

And similarly, ∠3 = ∠ 4 ( As they are bisector of ∠ BOQ)

From the straight line AB we know that angle made on straight line makes 180°

Therefore, ∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 = 180°

2 ( ∠ 2 + ∠ 4) = 180°

∠ 2 + ∠ 4 = 90°

Hence, Proved.

By axiom 1 we get,

Things which are equal to the same thing are equal to one another.

Here, 60 and b both are equal to a, hence b = 60.

One can draw an infinite number of lines through a given point on a plane.

Exactly one line can be drawn passing through two given points in a plane.

If two angles are supplementary, then their sum is = 180°.

Let, ∠x and ∠y are two supplementary angles.

∴ ∠x + ∠y = 180°

Let, the angle = 5x and the supplement = x

∴5x + x = 180°

⇒ 6x = 180°

⇒ x = 30°

∴ Measure of the angle = 5 × 30°= 150°

A pair of angles is complementary if the sum of their measures adds up to 90°s.

A pair of angles supplementary if the sum of their measures adds up to 180°s.

Δ ABC is a right angle. Triangle (where ∠ABC = 90°)

Here, ∠BAC + ∠ACB =90°

∴ ∠BAC and ∠ACB are complementary angle.

In the 2nd figure,

∠XYO + ∠OYZ = 180°

∴ ∠XYO and ∠OYZ are supplementary angles.

A plane is determined by 3 non-collinear points.

Two angles are adjacent where they have a common side and a common vertex and they do not overlap.

Here, ∠ROS and ∠SOT are adjacent to each other.

They have a common side SO and a common vertex O.

According to the problem,

⇒ AD = BC

⇒ AB – BD = AB – AC

⇒ AC = BD

Let, ∠BOX= ∠DOX = k°

According the figure,

∠BOD + ∠AOD = 180°

According the problem,

∠XOY =90° ⇒ ∠XOD + ∠YOD = 90° ⇒ ∠YOD = 90° -k°

∴ ∠BOX + ∠AOY = 90°

∠AOY = 90°- k°

∴ ∠AOY = ∠YOD

OY bisects ∠AOD

We know, ∠ALP + ∠BLP = 180°

According to problem,

∠ALR = ∠RLP = ∠ALD/2

Similarly,

∠BLS = ∠SLP = ∠BLP/2

∴ ∠RLP + ∠SLP = ∠ALD/2 + ∠BLD/2

⇒ ∠RLS = (∠ALP + ∠BLP)/2 = 180° = 90°

∴ ∠LRS + ∠RSL = 180° - ∠RLS

∠LRS + ∠RSL =180° - 90°

=90°

We know, ∠CVP = 70°,

∴∠UVW = 70° [∵ Vertically opposite angles]

Again,

⇒ ∠UWV + ∠DWU = 180°

⇒ ∠UWV = 180° - 110°

⇒ ∠UWV = 70°

In ΔUVW,

⇒ ∠UVW + ∠UWV + ∠VUW = 180°

⇒ 70° + 70° + ∠VUW = 180°

⇒ ∠VUW = 180° - 140° = 40°

∴ ∠QUS = ∠VUW = 40° [Vertically opposite angle]

Hour’s hand rotates 360° in = 12 hour = 720 min

Minute’s hand rotates 360° in = 1 hour = 60 min

i) At, 1.40hrs

Time past after 0.00 hrs = 1 hr + 40 min = 100 min

In 100 min hour’s hand rotate =

In 100 min⁡minute’s hand rotate =

∴ Difference between the hands,

= 240° - 50°

= 190°

ii) At 2.15 hrs,

Time past after 0.00 hrs = 2 hr + 15 min = 135 min

In 135 min hour’s hand rotate =

In 135 min minute’s hand rotate =

∴ Difference between the hands,

= 90° - 67.5°

= 22.5°

At 4.24 hrs,

Time past after 12.00 hrs = 4 hr + 24 min = 264 min

In 264 min hour’s hand rotate =

In 264 min minute’s hand rotate =

∴ Difference between the hands,

= 144° - 132°

= 12°

AD + BD

⇒ (AB + BD) + BD

⇒ AB + 2BD

⇒ 2BC + 2BD

⇒ 2(BC + BD)

⇒ 2CD

OX is extended to OY

OX is bisector of ∠BOD

∴ ∠DOX = ∠BOX

From figure,

∠DOX =∠COY

And ∠BOX = ∠AOY

∴ ∠COY = ∠AOY

∴OY intersect ∠AOC

⇒ ∠XOA + ∠XOB

⇒ ∠XOA (∠XOA + ∠AOB)

⇒ 2∠XOA + 2∠AOC

⇒ 2(∠XOA + ∠AOC)

⇒ 2∠XOC

⇒ ∠AOX - ∠XOB

⇒ (∠AOC + ∠COX) – (∠BOC - ∠COX)

⇒ ∠AOC - ∠BOC + 2∠COX

⇒ 2∠COX (∵ ∠AOC = ∠BOC)

∠AOC and ∠COB are perpendicular to each other.

∴ ∠AOC = 90° and ∠BOC = 90°

∴ ∠AOB = ∠AOC + ∠BOC = 90° + 90° = 180°

∴ O ,A and B are Collinear.

⇒ ∠ABC =180° - ∠GBC =180° - ∠DHC …… (1)

And

⇒ ∠CDE = 180°- ∠HDC …… (2)

From (1) + (2) we get,

⇒ ∠ABC + ∠CDE = 180° - ∠DHC + 180°- ∠HDC

⇒ ∠ABC + ∠CDE = 360° - (∠DHC + ∠HDC)

⇒ ∠ABC + ∠CDE = 360° - (180° - ∠HCD) [from ΔDHC]

⇒ ∠ABC + ∠CDE = 180° + ∠DCB

⇒ ∠ABC – ∠DCB + ∠CDE = 180°

AB and CD are two parallel lines.

EF is the transversal, which cuts AB at G and CD at H.

Here, ∠AGE = ∠BGH [Vertically opposite angle]

∠AGE = ∠GHC [Corresponding angle]

∠GHC = ∠FHD [Vertically opposite angle]

∴ ∠AGE = ∠BGH = ∠GHC = ∠FHD

Similarly, ∠BGE = ∠AGH [Vertically opposite angle]

∠BGE = ∠GHD [Corresponding angle]

∠GHD = ∠FHC [Vertically opposite angle]

∴ ∠BGE = ∠AGH = ∠GHD = ∠FHC

So, we have only 2 distinct values.