Congruence is a the state of agreement. It meaning is taken from a Latin word congruō which means “I meet together, I agree”. There is a difference between congruence and similarity. Congruence, as opposed to approximation, is a relation which implies a species of equivalence. If their corresponding sides are equal in length, in which case their corresponding angles are equal in size then two triangles are called congruent. Let two triangle ABC and triangle PQR, the congruence relationship can be written as, $\Delta ABC \cong \Delta PQR$ |

The proofs of congruency is given below,

Let two triangle ABC and triangle PQR. They are in same size and same shape. Hence they called as congruent triangle.

So, we can written as

Let two triangle ABC and triangle PQR. They are in same size and same shape. Hence they called as congruent triangle.

So, we can written as

$\Delta ABC = \Delta PQR$

From the above we can say that when you place PQR on ABC there will be some changes. Theses changes are P will falls on A, Q will falls on B and R will falls on C and also PQ will be on AB , QR on BC and PR on AC.

Consider these two triangles are congruent, then angle and sides match one another are equal. Thus, We have these two congruent triangles.

Corresponding vertices of triangles are A and P, B and Q, C and R.

Corresponding sides of triangles are AB and PQ, BC and QR , AC and PR.

Corresponding angles of triangles are $\angle A$ and $\angle P$, $\angle B$ and $\angle Q$, $\angle C$ and $\angle R$.

From the above we can say the matching of vertices also matters in congruence triangles.

Thus,

Thus,

$A \cong P$, $B \cong Q$, $C \cong R$

It can be written as,

$ABC \cong PQR$

When we have two triangles, how can we tell if they're congruent? They may look the same, but you can be certain by using one of several triangle congruence postulates, such as SSS, SAS, AAA or ASA.

• First, there's the side-side-side postulate, or SSS.

• Second, there's the side-angle-side postulate, or SAS.

• Third, there's the angle-side-angle postulate, or ASA.

• Forth, there's the angle-angle-angle postulate, or AAA.

SSS Triangles Congruence is nothing but the side side side congruency . This is the criteria for congruency of two triangles. If all three sides of two triangles are equal to each other triangles side, then theses triangles are said to be SSS congruent.

Let in the below given figure, AB = CD and AD = BC. Prove that $\Delta ADC$ is congruent to $\Delta CBA$.

Let in the below given figure, AB = CD and AD = BC. Prove that $\Delta ADC$ is congruent to $\Delta CBA$.

From the figure we have $\Delta ADC$ and $\Delta CBA$,

AB = CD ( by data )

AD = BC ( by data)

AC = AC ( by common side)

Hence Using SSS congruence we get,

$\Delta ADC \cong \Delta CBA$

SSS Congruence is a postulate states. That If the vertices of one triangle are equal to vertices of the other triangle.

Definition of AAS congruence is that two triangles are congruent if any two angles and single side of the triangle are equal to the corresponding sides and angles of the other triangle.

Let the triangle HGO is the isosceles triangle. OP is the bisector to the base HG. Prove that HOP is congruent to GOP.

Given Two triangle HOP and POG

From the figure,

$\angle HOP = \angle POG$ (PH is bisector)

$\angle OPH = \angle OPG = 90^{\circ}$

HO = GO ($\Delta HGO$ is isosceles)

It satisfies AAS condition

so

$\Delta OPH \cong \Delta OPG$

Congruence of AAS states that the shape and size of the two triangles are similar. Symbolic representation of AAS Congruence of triangle is $\cong$ . The symbol (~) is equivalent shape and (=) is equivalent size.

**Example 1:**

Prove that AP = BQ (see the figure).

**Solution:**

Given $\angle BAQ = \angle ABP$ and AQ = BP

Prove: AP = BQ

From the above image,

$\Delta ABP$ and $\Delta ABQ$

BP = AQ (from given)

AB = AB (common)

$\angle ABP = \angle BAQ$ (from given)

$\angle ABP \equiv \angle BAQ$ (SAS)

Hence from above we get,

AP = BQ

**Example 2:**

Equilateral $\Delta ABD$ and $\Delta ACE$ are drawn on the sides of a $\Delta ABC$, prove that CD = BE.

**Solution:**

Given ABC is a triangle. ABD and ACE are equilateral

Prove: CD = BE

Join BE and CD

From the above image, in $\Delta ABD$

$\angle BAD$ = $\angle ABD$ = $\angle BDA$ = $60^{\circ}$

in $\Delta ACE$,

$\angle AEC$ = $\angle ACE$ = $\angle CAE$ = $60^{\circ}$

$\angle CAD = \angle BAC + \angle DAB$

$\angle CAD = \angle BAC + 60^{\circ}$ (from given)

$\angle CAD = \angle BAC + \angle CAE$

$\angle CAD = \angle BAE$

$\Delta ACD and \Delta ABE$

AD = AB

AC = AE

$\Delta ACD\ \cong\ \Delta AEB$ (SAS)

Hence from above we get,

$CD = BE$