Congruent Triangles

Congruent Triangles

Congruent Triangles

Two triangles are said to be congruent if they are exactly same by their shapes and sizes. Two triangles of the same shapes is mean three angles of one triangle are equal to three angles of the other triangle and the triangles of same size mean the length of 3 sides of one triangle are equal to the three corresponding sides of the other triangle respectively. The symbol for congruency is .
triangles DEF and MNP
For example: In the given triangles DEF and MNP, DE = MN, EF = NP, FD = PM. Also, D = M, E = N and F = P. So the triangles DEF and MNP are congruent. It is written as ΔDEF ΔMNP.

Conditions for Congruency

The following conditions are necessary for the congruency of two triangles.

Condition - 1. Side-Side-Side (SSS) axiom

When three sides of a triangle are equal to three corresponding sides of another triangle, they are congruent by SSS axiom.
Side-Side-Side (SSS) axiom
In the given triangles, MN = DE (S), NP = EF (S), PM = FD (S). Then, ΔMNP ΔDEF by SSS axiom.

Condition - 2. Side-Angle-Side (SAS) axiom

When two sides of a triangle and the angle made by these sides are respectively equal to the two corresponding sides and the angle made by these sides of another triangle, then the triangles are congruent by SAS axiom.
Side-Angle-Side (SAS) axiom
In the given triangles, MN = DE (S), N = E (A), NP = EF (S). Then, ΔMNP ΔDEF by SAS axiom.

Condition - 3. Angle-Side-Angle (ASA) axiom

When two angles and a side included by these angles of a triangle are respectively equal to the two corresponding angles and a side included by these angles of another triangle, then the triangles are congruent by ASA axiom.
Angle-Side-Angle (ASA) axiom
In the given triangles, N = E (A), NP = EF (S), P = F (A). Then, ΔMNP ΔDEF by ASA axiom.

Condition - 4. Right angle-Hypotenuse-Side (RHS) axiom

In two right angled triangles hypotenuse and a side of a triangle are equal to the hypotenuse and a side of another triangle, then the triangles are congruent by RHS axiom.
Right angle-Hypotenuse-Side (RHS) axiom
In the given triangles, N = E = 90° (R), MP = DF (H), NP = EF (S). Then, ΔMNP ΔDEF by RHS axiom.

Condition - 5. Angle-Angle-Side (AAS) axiom

When two angles and non included side of a triangle are equal to the two angles and non included side of another triangle, then the triangles are congruent by AAS axiom.
Angle-Angle-Side (AAS) axiom
In the given triangles, M = D (A), P = F (A), NP = EF (S). Then, ΔMNP ΔDEF by AAS axiom.


Workout Examples

Example 1: In the given figure, state the condition of congruency and write their corresponding sides and angles.
Example 1: ΔPQR and ΔABC
Solution: Here,
                      In ΔPQR and ΔABC,
       i.        Q = R (A) ----------> both 60°
       ii.      QR = BC (S) ----------> both 5cm
       iii.    R = C (A) ----------> both 70°
  ΔPQR  ΔABC -------------> by ASA axiom
  PQ = AB, PR = AC ---------> corresponding sides of congruent triangles
  P = A ----------> corresponding angles of congruent triangles


Example 2: In the given figure, BP = PC, AP BC, prove that ΔABP ΔAPC, also find the value of a.
Example 2: ΔABP and ΔAPC
Solution: Here,
                      In ΔABP and ΔAPC,
       i.        BP = PC (S) ----------> given
       ii.      APB = APC (A) ----------> both right angles
       iii.    AP = AP (S) ----------> common side of both triangles
  ΔABP  ΔAPC -------------> by SAS axiom
  AB = AC ----------> corresponding sides of congruent triangles
                or,     2a + 3 = 5
                or,     2a = 5 – 3
                or,     2a = 2
                or,     a = 1


Example 3: In the given figure, AP = DP, CP = BP, PAC = 30° and PDB = 40° then,
a.       Prove that ΔAPC ΔDPB
b.       Find the values of a, b and y.
Example 3: ΔAPC and ΔBPD
Solution: Here,
                      In ΔAPC and ΔBPD,
       i.        AP = PD (S) ----------> given
       ii.      APC = DPB (A) ----------> vertically opposite angles
       iii.    PC = PB (S) ----------> given
  ΔAPC  ΔDPB -------------> by SAS axiom
  a  = 40° ----------> corresponding angles of congruent triangles
  b  = 30° ----------> corresponding angles of congruent triangles
  AC = DB ---------> corresponding sides of congruent triangles
                or,     y + 3 = 2y
                or,     3 = 2y – y
                or,     y = 3


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