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 ≅.
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.
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.
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.
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.
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.
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.
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.
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.
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
You can comment
your questions or problems regarding congruent triangles here.
No comments: