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.

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)

Example 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 $\triangle AEB \cong \triangle ADC$

$\\$

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.

Example 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 $CB = CD$

Example 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.

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.

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.

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.

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.

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 $BC = EF$

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 $\angle DEF = 90^o$

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 $\triangle ABC$

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: $AC > AB$

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:

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.

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 $\triangle DEB$

$AC = BD$ (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.