Revision: Geometry Mathematics SSLC (English Medium) Class 9 Tamil Nadu Board of Secondary Education

 The two rays forming the angle are called the arms or sides of the angle.

An angle consists of two rays that originate from a single initial point.  An angle is represented by the symbol ∠.

Angle: An angle consists of two rays that originate from a single initial point.  An angle is represented by the symbol ∠.

Arms of an Angle: The two rays forming the angle are called the arms or sides of the angle.

Vertex: The vertex of the angle is the common initial point of two rays.
Example:

The vertex of the angle is the common initial point of two rays.

Complementary angles: When the sum of the measures of two angles is 90°, the angles are called complementary angles. Example, 30° + 60° = 90°.

Supplementary angles: When the sum of the measures of two angles is 180° are called supplementary angles. Example, 60° + 120° = 180°.

Linear pair: A linear pair is a pair of adjacent angles whose non-common sides are opposite rays.

A plane figure bounded by three lines in a plane is called a triangle. The figure below represents a ΔABC, with AB, AC andBC as the three line segments.

A triangle (denoted by the symbol △) is the simplest closed shape in geometry. It is a two-dimensional figure made by connecting three points that do not lie on the same straight line (non-collinear).

SAS Congruence criterion: If under a correspondence, two sides and the angle included between them of a triangle are equal to two corresponding sides and the angle included between them of another triangle, then the triangles are congruent.

ASA Congruence criterion: If under a correspondence, two angles and the included side of a triangle are equal to two corresponding angles and the included side of another triangle, then the triangles are congruent.

RHS Congruence criterion: If under a correspondence, the hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right-angled triangle, then the triangles are congruent.

Parallelogram: A parallelogram is a quadrilateral whose opposite sides are parallel.

Rhombus: A rhombus is a quadrilateral with four equal-length sides and opposite sides parallel to each other.

• Trapezium: A trapezium is a quadrilateral where only two sides are parallel to each other.

• Isosceles trapezium: If the non-parallel sides of a trapezium are of equal length, we call it an isosceles trapezium.

Kite: A kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other.

A circle is a closed curve where all points on the boundary (called the circumference) are at the same distance from a fixed point inside it.

• The fixed point inside the circle is called the center (O)

The radius is a straight line segment that connects the center of the circle to any point on its circumference.

Characteristics:

• Symbol: Usually represented as r

• All radii of a circle have the same length

• A circle has infinite radii (one to every point on the circumference)

• The radius is always half the diameter

• Radius = `"Diameter"/"2"`

 The diameter is a straight line segment that passes through the center of the circle and has both endpoints on the circumference.

Characteristics:

• The diameter passes through the center

• A circle has infinite diameters

• The diameter is the longest possible chord of a circle

• The diameter is twice the radius

• Diameter = 2 × Radius and

A chord is a straight line segment that connects any two points on the circumference of the circle.

Characteristics:

• A circle has infinite chords

• The diameter is the longest chord in any circle

• Chords closer to the centre are longer than chords farther from the center

The line segment, joining any two points on the circumference of the circle, is called a chord. 

A circumcircle is a circle that passes through all three vertices of a triangle. The three vertices lie on the boundary of the circle.

The circumcenter is the center point of the circumcircle. It is the unique point where all three perpendicular bisectors of the triangle's sides meet.

• The circumcenter is equidistant from all three vertices of the triangle.

The circumradius is the radius of the circumcircle. It is the distance from the circumcenter to any vertex of the triangle.

The In-Circle of a triangle is the largest possible circle that can be drawn inside the triangle such that it just touches (is tangent to) all three sides.

The point where all three angle bisectors of a triangle meet. This point is the center of the In-Circle.

The perpendicular distance from the Incenter (I) to any of the three sides. This distance is the radius of the In-Circle.

Given: A(ΔPQR) in which QX is the bisector of ∠Q and RX is the bisector of ∠R.

XS ⊥ QR and XT ⊥  PQ.

We need to prove that:

1. ΔXTQ ≅ ΔXSQ.
2. PX bisects angle P.

Construction: Draw XZ ⊥ PR and join PX.

i. In ΔXTQ and ΔXSQ,

∠QTX = ∠QSX = 90°  ...[XS ⊥ QR and XT ⊥  PQ]

∠TQX = ∠SQX    ...[QX is bisector of ∠Q]

QX = QX    ...[Common]

∴ By Angle-Side-Angle Criterion of congruence,

ΔXTQ ≅ ΔXSQ

ii. The corresponding parts of the congruent triangles are congruent.

∴ XT = XS   ...[c.p.c.t.]

In ΔXSR and ΔXRZ

∠XSR = ∠XZR = 90°   ...[XS ⊥ QR and ∠XSR = 90°]

∠XRS = ∠ZRX      ...[RX is bisector of ∠R]

RX = RX    ....[Common]

∴ By Angle-Angle-Side criterion of congruence,

ΔXSR ≅ ΔXRZ

The corresponding parts of the congruent triangles are congruent.

∴ XS = XT    ...[c.p.c.t.] 

From (1) and (2)

XT = XZ                    

In ΔXTP and ΔPZX

∠XTP = ∠XZP = 90°    ....[Given]

XP = XP         ....[Common]

XT = XZ               

∴ By Right angle-Hypotenuse-side criterion of congruence,

ΔXTP ≅ ΔPZX

The corresponding parts of the congruent triangles are
congruent.

∴ ∠TPX = ∠ZPX    ...[c.p.c.t.]

∴ PX bisects ∠P.

Theorem (ASA congruence rule) :  Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle. 
Proof : We are given two triangles ABC and DEF in which: 
∠ B = ∠ E, ∠ C = ∠ F 
and BC = EF 
We need to prove that   ∆ ABC ≅ ∆ DEF
For proving the congruence of the two triangles see that three cases arise.

Case (i) : Let AB = DE in following fig.

You may observe that
AB = DE (Assumed) 
∠ B = ∠ E (Given)
BC = EF (Given) 
So, ∆ ABC ≅ ∆ DEF   (By SAS rule)

Case (ii) : 
Let if possible AB > DE. So, we can take a point P on AB such that PB = DE. Now consider ∆ PBC and ∆ DEF  in following fig. 

Observe that in ∆ PBC and ∆ DEF,
PB = DE (By construction) 
∠ B = ∠ E (Given)
BC = EF (Given)
So, we can conclude that: 
∆ PBC ≅ ∆ DEF, by the SAS axiom for congruence.
Since the triangles are congruent, their corresponding parts will be equal. 
So, ∠ PCB = ∠ DFE
But, we are given that
∠ ACB = ∠ DFE So, ∠ ACB = ∠ PCB
This is possible only if P coincides with A.
or, BA = ED 
So, ∆ ABC ≅ ∆ DEF   (by SAS axiom)

Case (iii) :  If AB < DE, we can choose a point M on DE such that ME = AB and repeating the arguments as given in Case (ii),  we can conclude that AB = DE and so, ∆ ABC ≅ ∆ DEF. 
You know that the sum of the three angles of a triangle is 180°. So if two pairs of angles are equal, the third pair is also equal (180° – sum of equal angles).
So, two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal. We may call it as the AAS Congruence Rule.

Theorem: A diagonal of a parallelogram divides it into two congruent triangles.
Proof :  Let ABCD be a parallelogram and AC be a diagonal in following fig. 

Observe that the diagonal AC divides parallelogram ABCD into two triangles, namely, ∆ ABC and ∆ CDA. We need to prove that these triangles are congruent.
In ∆ ABC and ∆ CDA, note that BC || AD and AC is a transversal.
So, ∠ BCA = ∠ DAC (Pair of alternate angles)
Also, AB || DC and AC is a transversal.
So, ∠ BAC = ∠ DCA (Pair of alternate angles)
and AC = CA (Common)
So, ∆ ABC ≅ ∆ CDA (ASA rule)
or, diagonal AC divides parallelogram ABCD into two congruent triangles ABC and CDA. 
Now, measure the opposite sides of parallelogram ABCD. 
You will find that AB = DC and AD = BC.

The opposite sides of a parallelogram are of equal length.

Given: ABCD is a parallelogram.

To Prove: AB = DC and BC = AD.

Construction: Draw any one diagonal, say `bar(AC)`.

Proof:

Consider a parallelogram ABCD,

In triangles ΔABC and ΔADC,

∠ 1 = ∠2, ∠ 3 = ∠ 4             .....(Pair of alternate angle)
and `bar(AC)` is common side.

Side AC = Side AC              .....(common side)
∠ 1 ≅ ∠2                             .....(Pair of alternate angle)
∠ 3 ≅ ∠ 4                            .....(Pair of alternate angle)

by ASA congruency condition,
∆ ABC ≅ ∆ CDA

This gives AB = DC and BC = AD.

Hence Proved.

Theorem: Parallelograms on the same base and between the same parallels are equal in area.
Proof : Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF and DC are given in following fig. 

We need to prove that ar (ABCD) = ar (EFCD).
In ∆ ADE and ∆ BCF, 
∠ DAE = ∠ CBF (Corresponding angles from AD || BC and transversal AF)   (1)
∠ AED = ∠ BFC (Corresponding angles from ED || FC and transversal AF) (2)
Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3)
Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)
So, ∆ ADE ≅ ∆ BCF         [By ASA rule, using (1), (3), and (4)]
Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)
Now, ar (ABCD) = ar (ADE) + ar (EDCB)
= ar (BCF) + ar (EDCB)         [From(5)]
= ar (EFCD)
So, parallelograms ABCD and EFCD are equal in area.

Theorem : Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Proof: Now, suppose ABCD is a parallelogram whose one of the diagonals is AC (see Fig.). 

Let AN ⊥ DC. Note that
∆ ADC ≅ ∆ CBA 
So, ar (ADC) = ar (CBA) 
Therefore, ar (ADC) = `1/2` ar (ABCD) 
=`1/2` (DC × AN)
So, area of ∆ ADC = `1/2` × base DC × corresponding altitude AN
In other words, area of a triangle is half the product of its base (or any side) and the corresponding altitude (or height). 
The two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.

We have to prove that

AQ = `1/2` (perimeter of ΔABC)

Perimeter of ΔABC = AB + BC + CA

= AB + BP + PC + CA

= AB + BQ + CR + CA

(∵ Length of tangents from an external point to a circle are equal ∴ BP = BQ and PC = CR)

= AQ + AR  ...(∵ AB + BQ = AQ and CR + CA = AR)

= AQ + AQ  ...(∵ Length of tangents from an external point are equal)

= 2AQ

⇒ AQ = `1/2` (Perimeter of ΔABC)

Hence proved.

Theorem: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

If in a circle with centre O , arc PQ of a circle subtends ∠POQ at centre and arc PQ subtends ∠PAQ on the remaining part of the circle than  ∠POQ = 2 PAQ.

Theorem: Angles in the same segment of a circle are equal.  
Suppose we join points P and Q and form a chord PQ in the above figures. Then ∠ PAQ is also called the angle formed in the segment PAQP.

∠ POQ = 2 ∠ PCQ = 2 ∠ PAQ 
Therefore, ∠ PCQ = ∠ PAQ.
 Here ∠PAQ is an angle in the segment, which is a semicircle. Also, ∠ PAQ = `1/2`
∠POQ = `1/2 xx = 180° =90° ` 

Theorem:   If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).
In following fig. AB is a line segment, which subtends equal angles at two points C and D. That is ∠ ACB = ∠ ADB 

To show that the points A, B, C and D lie on a circle. let us draw a circle through the points A, C and B. Suppose it does not pass through the point D. Then it will intersect AD (or extended AD) at a point, say E .
If points A, C, E and B lie on a circle, 
∠ ACB = ∠ AEB 
But it is given that ∠ ACB = ∠ ADB. 
Therefore, ∠ AEB = ∠ ADB. 
This is not possible unless E coincides with D. 
Similarly, E′ should also coincide with D.

Given:

O is the circumcenter of triangle ABC.

D is the foot of the perpendicular from O to BC.

So OD ⟂ BC.

We have to prove that   ∠BOD = ∠A

OB = OC = OA

Since O is the circumcenter, all these are radii of the circumcircle of triangle ABC.

OD ⟂ BC ⇒ D is midpoint of BC

In a circle, the perpendicular drawn from the center to any chord bisects that chord.

BC is a chord and OD ⟂ BC; hence,

BD = DC

D is the midpoint of BC.

OD bisects ∠BOC

OB = OC (radii)

BD = DC (from Step 2)

O and D lie on perpendicular bisector of BC

Therefore, OD is the perpendicular bisector of chord BC; hence, it bisects the angle at the centre subtended by BC.

∠BOD = ∠COD

Using the Central Angle Theorem

Chord BC subtends:

At center: ∠BOC

At circumference (on ΔABC): ∠A

The angle made at the center is twice the angle made at the circumference by the same chord.

∠BOC = 2∠A     ...[OD bisects ∠BOC]

Statement:
The angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle.

Result:

∠AOB = 2∠ACB

Short Proof (Idea):

• Join the centre to the points on the circle.

• Radii form isosceles triangles.

• Angle at centre = sum of angles at the circle.

• Hence, the angle at the centre is double the angle at the circle.

Statement:
Angles in the same segment of a circle are equal.

Result:

∠ACB = ∠ADB

Short Proof (Idea):

• Both angles stand on the same arc.

• The angle at the centre is double each of them.

• Hence, both angles are equal.

Statement:
The angle in a semicircle is a right angle.

Result:

∠ACB = 90^∘

Short Proof (Idea):

• Diameter subtends an angle of 180° at the centre.

• The angle at the circle is half of it.

• Therefore, angle = 90°.

Statement:
If an arc of a circle subtends a right angle at any point on the remaining part of the circle, then the arc is a semicircle.

Result:
Arc AB is a semicircle
(or AB is a diameter)

Short Proof (Idea):

• Given angle at the circle ∠ACB = 90°.

• The angle at the centre is double the angle at the circle.

• Therefore, ∠AOB = 2 × 90° = 180°.

• Hence, A, O, and B lie on a straight line, so AB is a diameter.

• Therefore, arc AB is a semicircle.

Let ABCD be a parallelogram inscribed in a circle.

Now, ∠BAD + ∠BCD

(Opposite angles of a parallelogram are equal.)

And ∠BAD + ∠BCD = 180°

(A pair of opposite angles in a cyclic quadrilateral are supplementary.)

∠BAD + ∠BCD = `(180^circ)/2` = 90°

The other two angles are 90°, and the opposite pair of sides are equal.

∴ ABCD is a rectangle.

Statement:
The sum of the opposite angles of a cyclic quadrilateral is 180°.

Short Proof (Idea):

• Let ABCD be a cyclic quadrilateral.

• Arc ABC subtends an angle ∠ADC at the circle and ∠AOC at the centre.

• The angle at the centre is double the angle at the circle.

  ∠ADC=`1/2`∠AOC

• Similarly, the other arc subtends:

  ∠ABC = `1/2`(reflex ∠AOC)

• The sum of angles around the centre is 360°.

• Therefore,

  ∠ADC + ∠ABC = `1/2`(360°) = 180^∘

Conclusion:
Hence, the opposite angles of a cyclic quadrilateral are supplementary.

Given: `square`ABCD is a cyclic quadrilateral.

To prove: ∠BAD + ∠BCD = 180º 

∠ABC + ∠ADC = 180º

Proof: Arc BCD is intercepted by the inscribed ∠BAD.

∠BAD = `1/2` m(arc BCD)     ...(i) [Inscribed angle theorem]

Arc BAD is intercepted by the inscribed ∠BCD.

∴ ∠BCD = `1/2` m(arc DAB)      ...(ii) [Inscribed angle theorem]

From (1) and (2) we get

∠BAD + ∠BCD = `1/2` [m(arc BCD) + m(arc DAB)]

∴ (∠BAD + ∠BCD) = `1/2 xx 360^circ`    ...[Completed circle]

= 180°

Again, as the sum of the measures of angles of a quadrilateral is 360°

∴ ∠ADC + ∠ABC = 360° – [∠BAD + ∠BCD]

= 360° – 180°

= 180°

Hence, the opposite angles of a cyclic quadrilateral are supplementary.

Consider a ΔABC.

Two circles are drawn while taking AB and AC as the diameter.

Let them intersect each other at D, and let D not lie on BC.

Join AD.

∠ADB = 90°...(Angle subtended by semi-circle)

∠ADC = 90°   ...(Angle subtended by semi-circle)

∠BDC = ∠ADB + ∠ADC = 90° + 90° = 180°

Therefore, BDC is a straight line, and hence, our assumption was wrong.

Thus, point D lies on the third side BC of ΔABC.

It is given that BE is the bisector of ∠B.

∴ ∠ABE = ∠B/2

However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)

⇒ ∠ADE = ∠B/2

Similarly, ∠ACF = ∠ADF = ∠C/2         (Angle in the same segment for chord AF)

∠D = ∠ADE + ∠ADF

`= (angleB)/2 + (angleC)/2`

`= 1/2(angleB + angleC)`

`= 1/2(180^@ - angleA)`

`= 90^@ - 1/2angleA`

Similarly, it can be proved that

`angleE = 90^@ - 1/2angleB`

`angleF = 90^@ - 1/2angleC`

Let two circles having their centres as O and O’ intersect each other at point A and B respectively. Let us join OO’.

In ΔAOO’ and BOO’,

OA = OB   ...(Radius of circle 1)

O’A = O’B   ...(Radius of circle 2)

OO’ = OO’   ...(Common)

ΔAOO’ ≅ ΔBOO’   ...(By SSS congruence rule)

∠OAO’ = ∠OBO’   ...(By CPCT)

Therefore, line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Statement:
If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.

Short Proof (Idea):

• Given, ∠B + ∠D = 180°.

• Draw a circle through three vertices of the quadrilateral.

• If the fourth vertex does not lie on the circle, an exterior angle becomes equal to its interior opposite angle, which is not possible.

• Hence, the fourth vertex must lie on the same circle.

Conclusion:
Therefore, ABCD is a cyclic quadrilateral.

Statement:

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Short Proof (Idea):

• In a cyclic quadrilateral, the sum of opposite angles is 180°.

• The exterior angle and the adjacent interior angle form a straight line, so their sum is 180°.

• Since both are supplementary to the same angle, they are equal.

Conclusion:

Therefore, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

In ΔABC,

∠ABC + ∠BCA + ∠CAB = 180°  ...(Angle sum property of a triangle)

⇒ 90° + ∠BCA + ∠CAB = 180°

⇒ ∠BCA + ∠CAB = 90°    ...(1)

In ΔADC,

∠CDA + ∠ACD + ∠DAC = 180°   ...(Angle sum property of a triangle)

⇒ 90° + ∠ACD + ∠DAC = 180°

⇒ ∠ACD + ∠DAC = 90°   ...(2)

Adding equations (1) and (2), we obtain

∠BCA + ∠CAB + ∠ACD + ∠DAC = 180°

⇒ (∠BCA + ∠ACD) + (∠CAB + ∠DAC) = 180°

∠BCD + ∠DAB = 180°     ...(3)

However, it is given that

∠B + ∠D = 90° + 90° = 180°    ...(4)

From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180°. Therefore, it is a cyclic quadrilateral.

Consider chord CD.

∠CAD = ∠CBD   ...(Angles in the same segment)

• AA / AAA → two angles equal

• SAS → included angle equal + sides proportional

• SSS → all sides proportional

• Diameter → Longest chord of a circle

• Perpendicular from centre to a chord → Bisects the chord

• Line joining centre to midpoint of a chord → Perpendicular to the chord

• The greater the chord → Nearer to the centre

• The smaller the chord, → farther from the centre

• Equal chords → Equidistant from the centre

• Chords equidistant from centre → Equal in length

• Only one circle passes through three non-collinear points

1. Arc Definition: An arc is a curved portion of a circle's circumference between two points.

2. Two Types: Minor arc (< 180°) and Major arc (> 180°).

3. Semicircle: When the arc angle is exactly 180°, it's called a semicircle.

4. Complete Circle: Minor arc + Major arc = 360° (complete circumference).