In this chapter we will learn about the concept of “Congruence” of triangles.

In simple, any two geometrical figures, exactly of the same size and shape are congruent.

Congruence of a line segment: Two line segment AB and CD are congruent if and only if their lengths are equal. i.e. if AB = CD

Congruence of angles: Two angles are congruent if and only if their measures are equal. i.e. \angle ABC = \angle DEF

Congruence of triangles:

Consider two \triangle ABC and \triangle DEF (as shown in the figure)

Two triangles are congruent if and only if one of them can be made to superpose on the other so as to cover it exactly.

Two triangles are congruent if and only if there exists a correspondence between their vertices such that the corresponding sides and the corresponding angles of the two triangles are equal. This means that if:

AB = DE , BC = EF \ and \  CA = FD and

\angle A = \angle D, \angle B = \angle E

and \angle C = \angle F

\Rightarrow \triangle ABC \cong \triangle DEF

Also if \triangle ABC \cong \triangle DEF   and \triangle ABC \cong \triangle PQR \Rightarrow \triangle DEF \cong \triangle PQR

 

Theorems of Congruence

Theorem 1: S.A.S (Side Angle Side): Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding sides and the included angle of the other triangle.Slide1

Example 1: In the adjoining figure, O is the mid point of AB and CD . Prove that (i) \triangle AOC \cong \triangle BOD (ii) AC = BD (iii) AC \parallel BD

Solution: Consider \triangle AOC and  \triangle BOD

AO = OB (Given)

CO = OB (Given)

\angle AOB = \angle DOC (vertically opposite angle)

Therefore  \triangle AOC \cong \triangle BOD (By S.A.S theorem)

Since the triangles are congruent, AC = BD (corresponding sides of congruent triangles are equal)

Also since the two triangles are congruent, \angle ACO = \angle ODB (corresponding angles are also equal). Hence AC \parallel BD (alternate angles are equal)

2018-07-12_9-19-05Example 2: In the adjoining figure, it is given that AE = AD and BD = CE . Prove that \triangle AEB \cong \triangle ADC .

Solution: BF is common segment of AF and CF

AE = AD and CE = BD

Therefore AE + CE = AD + BD \Rightarrow AC = AB

Now consider \triangle AEB and  \triangle ADC

AE = AD (given)

AC = AB (Proved above)

\angle A is common

Therefore \triangle AEB \cong \triangle ADC

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Theorem 2: A.S.A (Angle Side Angle): Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

2018-07-14_18-05-28Example 3: In the adjoining triangle, diagonal AC   is a quadrilateral ABCD bisects the \angle A and \angle C . Prove that AB = AD and CB = CD .

Solution:

Given: \angle BAC = \angle CAD and \angle BCA = \angle DCA

Consider \triangle ABC and \triangle ADC

\angle BAC = \angle CAD (given)

\angle BCA = \angle DCA (given)

and AC is common.

Therefore  \triangle ABC \cong \triangle ADC (A.S.A theorem)

Hence we can say that BA = DA and CB = CD

2018-07-14_18-29-17Example 4:   AB is a line segment, AC and BD are two equal line segments draw on opposite sides of line AB such that AC \parallel BD . If AB and CD intersect each other at O , prove that (i) \triangle AOC \cong \triangle BOD (ii) AB and CD bisect each other

Solution:

Given that AC \parallel BD and AB is the transversal.

Therefore \angle BAC = \angle ABD 

Similarly, AC \parallel BD and CD is the transversal.

Therefore \angle ACD = \angle BDC 

Now consider \triangle OAC and \triangle OBD

\angle BAC = \angle ABD

\angle ACD = \angle BDC

AC = BD (given)

Therefore  \triangle OAC \cong \triangle OBD

\Rightarrow AO = BO and OC = OD

 

Theorem 3.1 A.A.S (Angle Angle Side): If any two angles and a not included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.2018-07-24_8-07-31

Solution / Proof: 

In the given \triangle ABC and \triangle DEF , we have \angle B = \angle E , \angle A = \angle D and BC = EF .

Therefore \angle A + \angle B = \angle D + \angle E

\Rightarrow 180 -  ( \angle A + \angle B) = 180 - (\angle D + \angle E)

\Rightarrow \angle C = \angle F

Hence we see that with \angle B = \angle E, BC = EF and \angle C = \angle F

Hence \triangle ABC \cong \triangle DEF (A.S.A Theorem).

 

Theorem 3.2: If two angles of a triangle are equal, then sides opposite to them are also equal.2018-07-24_8-26-16

Solution:

Given: \angle B = \angle C

Draw the bisector of \angle A . Now consider \triangle ABD and \triangle ADC

We have \angle BAD = \angle DAC (angle bisector)

\angle B = \angle C

and AD is common or equal.

Hence \triangle ABD \cong \triangle ADC by (A.A.S theorem)

\Rightarrow AB = AC

 

Theorem 3.3: If the altitude of one vertex of a triangle bisects the opposite side, then the triangle is isosceles.2018-07-24_8-09-16

Solution:

In the \triangle ABC, AD is the altitude

\Rightarrow \angle ADB = \angle ADC = 90^o

BD = DC (given)

and AD is common.

Therefore \triangle ABD \cong \triangle ADC

\Rightarrow AB = AC

 

Theorem 3.4: In an isosceles triangle altitude from the vertex bisects the base.2018-07-24_8-09-45

Solution:

In \triangle ABC , given AB =AC \Rightarrow \angle B = \angle C

\angle ADB = \angle ADC = 90^o (Altitude)

And AD is common.

Therefore \triangle ADB \cong \triangle ADC (A.A.S theorem)

\Rightarrow BD = DC

 

Theorem 3.5: If the bisector of the vertical angle of a triangle bisects the base of the triangle, then the triangle is isosceles.2018-07-24_9-16-31

Solution:

Extend AD to E such that AD = DE

Given: BD = DC and \angle BAD = \angle CAD

Now, consider \triangle ABD and \triangle DEC

Therefore BD = DC (given) , AD = DE (constructed)

and \angle ADB = \angle CDE (vertically opposite angles)

Hence \triangle ABD \cong \triangle DEC  (S.A.S theorem)

Therefore AB = EC and \angle BAD = \angle CED

we know, \angle BAD = \angle CAD \Rightarrow \angle CAD = \angle CED

\Rightarrow AC = EC \Rightarrow AC = AB . Hence \triangle ABC is an isosceles triangle.

 

Theorem 4: S.S.S (Side Side Side):  Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.

Given two \triangle ABC and \triangle DEF such that AB = DE, AC = DF and BC = EF 2018-07-28_9-54-53

To Prove: \triangle ABC \cong \triangle DEF

Solution:

In \triangle DEF , draw EG such that \angle FEG = \angle ABC and EG = AB .

Join GF and GD

Consider, \triangle ABC and \triangle EFG We have

BC = EF (given)

AB = EG (by construction)

And \angle ABC = \angle FEG (by construction)

Therefore \triangle ABC \cong \triangle EFG (by S.A.S theorem)

\Rightarrow \angle BAC = \angle EGF and AC = FG

Since AB = DE and AB = GE \Rightarrow DF = FG

In \triangle DEG, DE = EG \Rightarrow \angle EDG = \angle EGD (angle opposite to the equal sides are equal)

Similarly, in In \triangle DFG, DF = FG \Rightarrow \angle FDG = \angle FGD (angle opposite to the equal sides are equal)

Hence \angle EDG + \angle FDG = \angle EGD + \angle FGD

\Rightarrow \angle EDF = \angle EGF

\Rightarrow \angle EGF = \angle BAC

\Rightarrow \angle BAC = \angle EDF

Now consider \triangle ABC and \triangle DEF

AB = DE (given)

\angle BAC = \angle EDF (just proved above)

AC = DF (given)

Therefore \triangle ABC \cong \triangle DEF (by S.A.S theorem). Hence proven.

 

Theorem 5: R.H.S (Right Angle Hypotenuse Side): Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the one side of the other triangle.

Given two right angled \triangle ABC and \triangle DEF . \angle ABC = 90^o and \angle DEF = 90^o 2018-07-28_9-56-01

Also given, AC = DF and BC = EF

To Prove: \triangle ABC \cong \triangle DEF

Solution: 

Consider \triangle ABC and \triangle GEF

AB = GE (by construction)

\angle ABC = \angle FEG = 90^o (by construction)

BC = EF (given)

Therefore  \triangle ABC \cong \triangle GEF (By S.A.S theorem)

\Rightarrow \angle BAC = \angle EGF

AC = GF and AC = DF \Rightarrow DF = GF

\Rightarrow \angle EDF = \angle EGF (angle opposite to the equal sides are equal)

Therefore \angle BAC = \angle EDF

In \triangle ABC and \triangle DEF , we have

\angle BAC = \angle EDF

\angle ABC =\angle DEF

\Rightarrow \angle BAC + \angle ABC = \angle EDF + \angle DEF

\Rightarrow 180^o - \angle BCA  = 180^o - \angle EFD

\Rightarrow  \angle BCA  = \angle EFD

Now consider \triangle ABC and \triangle DEF

BC = EF (given)

\angle BCA = \angle EFD (proved above)

AC = DF (given)

Therefore \triangle ABC \cong \triangle DEF (by S.A.S theorem)

 

Inequality Relations in a Triangle

Theorem 1: If two sides of a triangle are unequal, the longer side has greater angle opposite to it.

Given AC > AC in \triangle ABC 2018-07-28_11-13-22

Find a point D on AC such that

AD = AB \Rightarrow \angle ABD = \angle ADB … … … (i)

Consider \triangle BCD

We know \angle ADB = \angle DBC + \angle DCB (exterior angle is the sum of other two angles)

\Rightarrow \angle ADB > \angle DCB or \angle ADB > \angle ACB … … … … … (ii)

Therefore from (i) and (ii) we get

\angle ABD = \angle DBA and \angle ADB > \angle ACB

\Rightarrow \angle ABD > \angle ACB … … … … … (iii)

From \triangle ABC we get \angle ABC > \angle ABD … … … … … (iv)

From (iii) and (iv) we get

\angle ABC > \angle ACB . Hence proven.

 

Theorem 2: In a triangle the greater angle has the longer side opposite to it (converse of Theorem 1).

In \triangle ABC , given \angle ABC > \angle ACB

To Prove: AC > AB 2018-07-28_11-13-56

Solution: 

There are three possible cases (i) AC = AB (ii) AC < AB (iii) AC > AB

Case 1: If AC = AB , then \angle ABC = \angle ACB (angles opposite to equal sides are equal). But that is not true as it is  given that \angle ABC > \angle ACB

Case 2: When AC < AB \Rightarrow \angle ACB > \angle ABC (longer side has the greater angle opposite it). This not true either.

Case 3: Hence AC > AB is the only possibility that is left.

 

Theorem 3: The sum of any two sides of a triangle is greater than the third side.

Given \triangle ABC

To Prove: AB + AC > BC, AB + BC > AC and BC + AC > AB

Construction: Extend BA to D such that AC = AD Join CD

Solution: 2018-07-28_11-19-14

In \triangle ACD

AC = AD (by construction)

\Rightarrow \angle ADC = \angle ACD (angles opposite to equal sides are equal)

\Rightarrow \angle BCA + \angle ACD > \angle ADC

\Rightarrow \angle BCD > \angle ADC

\Rightarrow \angle BCD > \angle BDC

\Rightarrow BD > BC (since side opposite to the greater angle is larger)

\Rightarrow AB + AD > BC

\Rightarrow AB + AC > BC (since AC = AD by construction)

Similarly, you can prove that AB + BC > AC and BC + AC > AB

 

Theorem 4: Of all the line segment that can be drawn to a given line, from a point, not lying on it, the perpendicular line segment is the shortest.2018-07-28_11-19-32

Given that l is a straight line and point A is not lying on the line. AB \perp l and C is any other point B .

Prove: AB < AC

Solution: In \triangle ABC , we have \angle ABC = 90^o

\Rightarrow \angle ACB < 90^o (sum of the angle of a triangle is 180^o )

\Rightarrow \angle ACB < \angle ABC

\Rightarrow AB < AC . Hence proven.

 

Pythagoras Theorem

Theorem 1 (Pythagoras Theorem): In a a right triangle, the square of the hypotenuse is equal to the sum of the square of the two sides.. If in a \triangle ABC , if BC is the hypotenuse, then BC^2 = AC^2 + AB^2 .

Construction: Extend side AB to point D such that BD = CA = b . At D draw DE \perp AD and cut off DE = AB = c . Join CE .

Solution:

Consider \triangle ABC and \triangle DEB 2018-07-28_16-40-01

AC = BD (by construction)

\angle BAC = \angle EDB (right angles)

AB = DE (by construction)

\triangle ABC \cong \triangle DEB (by S.A.S theorem)

\Rightarrow BC = BE and \angle ACB = \angle DBE

We know \angle BAC + \angle ACB + \angle ABC = 180^o

\Rightarrow \angle ABC + \angle ACB = 90^o

\Rightarrow \angle ABC + \angle DBE = 90^o 

Also \angle ABC + \angle CBE + \angle DBE = 180^o

\Rightarrow 90^o+\angle CBE = 180

\Rightarrow \angle CBE = 90^o

That is \triangle CBE is a right angle triangle.

Therefore  at point B we have \angle CAB + \angle EDB = 180^o \Rightarrow CA \parallel ED

\Rightarrow ADEC is a trapezium

Now, area of trapezium ADEC = area of \triangle ABC + area of \triangle CBE + area of \triangle BDE

\Rightarrow \frac{1}{2} (CA + DE) =  \frac{1}{2} AB \times CA + \frac{1}{2} CB \times BE + \frac{1}{2} BD \times ED

\Rightarrow \frac{1}{2} (b+c)(b+c) = \frac{1}{2} bc + \frac{1}{2} a^2+ \frac{1}{2} bc

\Rightarrow (b+c)^2 = 2bc + a^2

\Rightarrow a^2 = b^2 +c^2

\Rightarrow BC^2 = AB^2 + AC^2

 

Theorem 2 (Converse of the Pythagoras Theorem): In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to this side is a right angle.

Given, \triangle ABC such that AC^2 = AB^2 + BC^2

To Prove: \angle ABC = 90^o

Construction: Draw another \triangle DEF such that AB = DE, BC = EF and \angle DEF = 90^o

Solution:

Since by construction, \triangle DEF is a right angled triangle, by Pythagoras theorem,

DF^2 = DE^2 + EF^2

\Rightarrow DF^2 = AB^2 + BC^2

\Rightarrow DF^2 = AC^2

\Rightarrow DF = AC

In \triangle ABC and \triangle DEF , we have AB = DE, BC = EF and AC = DF

Therefore \triangle ABC \cong \triangle DEF

\Rightarrow \angle ABC = \angle DEF = 90^o

Hence \triangle ABC is a right angled triangle.

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