Congruent Triangles

Congruent Triangles

Two triangles are said to be congruent triangles if they are exactly the same by their shapes and sizes.

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Two triangles of the same shape are meant three angles of one triangle are equal to the corresponding three angles of another triangle. And, two triangles of the same size are meant three sides of one triangle are equal to the corresponding three sides of another triangle.

The symbol is used for congruent triangles.

For example:

In the given triangles ABC and PQR, AB = PQ, BC = QR, AC = PR. Also, A = P, B = Q and C = R. So the triangles ABC and PQR are congruent. It is written as ΔABC ΔPQR.

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 Side-Side-Side condition. It is called SSS axiom.

In the given triangles, AB = PQ (S), BC = QR (S), AC = PR (S). Therefore, ΔABC ΔPQR by SSS axiom.

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

When Side-Angle-Side serially of a triangle are equal to the corresponding Side-Angle-Side serially of another triangle, then the triangles are congruent by Side-Angle-Side condition. It is called SAS axiom.

In the given triangles, AB = PQ (S), B = Q (A), BC = QR (S). Therefore, ΔABC ΔPQR by SAS axiom.

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

When Angle-Side-Angle serially of a triangle are equal to the corresponding Angle-Side-Angle serially of another triangle, then the triangles are congruent by Angle-Side-Angle condition. It is called ASA axiom.

In the given triangles, B = Q (A), BC = QR (S), C = R (A). Therefore, ΔABC ΔPQR by ASA axiom.

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

In two right-angled triangles, if the hypotenuse and a side of a triangle are equal to the hypotenuse and a side of another triangle, then the triangles are congruent by Right angle-Hypotenuse-Side condition. It is called RHS axiom.

In the given triangles, B = Q = 90° (R), AC = PR (H), BC = QR (S). Therefore, ΔABC ΔPQR by RHS axiom.

Condition - 5: Side-Angle-Angle (SAA) axiom

When Side-Angle-Angle serially of a triangle are equal to the corresponding Side-Angle-Angle serially of another triangle, then the triangles are congruent by Side-Angle-Angle condition. It is called SAA axiom.

In the given triangles, BC = QR (S), C = R (A), A = P (A). Therefore, ΔABC ΔPQR by SAA axiom.

Worked Out Examples

Example 1: In the given figure, state the condition of congruency and write their corresponding sides and angles.

Solution: Here,

In ΔPQR and ΔABC

i.       B = Q (A) -------> both 60°

ii.    BC = QR (S) -------> both 5cm

iii. C = R (A) -------> both 45°

ΔABC ΔPQR -------> by ASA axiom

AB = PQ, AC = PR -------> corresponding sides of congruent triangles

A = P -------> 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.

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  Ans.

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.

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  Ans.

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