# Side Side Side Congruence

Conditions for the SSS - Side Side Side congruence

Two triangles are said to be congruent if three sides of one triangle are respectively equal to the three sides of the other triangle.

Experiment to prove Congruence with SSS:

Draw ∆LMN with LM = 3 cm, LN = 4 cm, MN = 5 cm.

Also, draw another ∆XYZ with XY = 3cm, XZ = 4cm, YZ= 5cm.

We see that LM = XY, LN = XZ and MN = YZ.

Make a trace copy of ∆XYZ and try to make it cover ∆LMN with X on L, Y on M and Z on N.

We observe that: two triangles cover each other exactly.

Therefore ∆LMN ≅ ∆XYZ

Worked-out problems on side side side congruence triangles (SSS postulate):

1. LM = NO and LO = MN. Show that ∆ LON ≅ ∆ NML.

Solution:

In ∆LON and ∆NML

LM = NO   →  given

LO = MN   →  given

LN = NL   →   common

Therefore, ∆ LON ≅ ∆ NML, by side-side-side (SSS) congruence condition

2. In the given figure, apply SSS congruence condition and state the result in the symbolic form.

Solution:

In ∆LMN and ∆LON

LM = LO = 8.9cm

MN = NO = 4cm

LN = NL = 4.5 cm

Therefore, ∆LMN ≅ ∆LON, by side side side (SSS) congruence condition

3. In the adjoining figure, apply S-S-S congruence condition and state the result in the symbolic form.

Solution:

In ∆LNM and ∆OQP

LN = OQ = 3 cm

NM = PQ = 5cm

LM = PO = 8.5cm

Therefore, ∆LNM ≅ ∆OQP, by Side Side Side (SSS) congruence condition

4. ∆OLM and ∆NML have common base LM, LO = MN and OM = NL. Which of the following are true?

(i) ∆LMN ≅ ∆LMO

(ii)  ∆LMO ≅ ∆LNM

(iii) ∆LMO ≅ ∆MLN

Solution:

LO = MN and OM = NL   →    given

LM = LM    → common

Thus, ∆MLN ≅ ∆LMO, by SSS congruence condition

Therefore, statement (iii) is true. So, (i) and (ii) statements are false.

5. By Side Side Side congruence prove that 'Diagonal of the rhombus bisects each other at right angles'.

Solution: Diagonal LN and MP of the rhombus LMNP intersect each other at O.

It is required to prove that LM ⊥ NP and LO = ON and MO = OP.

Proof: LMNP is a rhombus.

Therefore, LMNP is a parallelogram.

Therefore, LO = ON and MO = OP.

In ∆LOP and ∆LOM; LP = LM, [Since, sides of a rhombus are equal]

Side LO is common

PO = OM, [Since diagonal of a parallelogram bisects each other]

Therefore, ∆LOP ≅ ∆LOM, [by SSS congruence condition]

But, ∠LOP + ∠MOL = 2 rt. angle

Therefore, 2∠LOP = 2 rt. angle

or, ∠LOP = 1 rt. angle

Therefore, LO ⊥ MP

i.e., LN ⊥ MP (Proved)

[Note: Diagonals of a square are perpendicular to each other]

6. In a quadrilateral LMNP, LM = LP and MN = NP.

Prove that LN ⊥ MP and MO = OP [O is the point of intersection of MP and LN]

Proof:

In ∆LMN and ∆LPN,

LM = LP,

MN = NP,

LN = NL

Therefore, ∆LMN ≅ ∆LPN, [by SSS congruence condition]

Therefore, ∠MLN = ∠PLN -------- (i)

Now in ∆LMO and ∆LPO,

LM = LP;

LO is common and

∠MLO = ∠PLO

∆LMO ≅ ∆LPO, [by SAS congruence condition]

Therefore, ∠LOM = ∠LOP and

MO = OP, [Proved]

But ∠LOM + ∠LOP = 2 rt. angles.

Therefore, ∠LOM = ∠LOP = 1 rt. angles.

Therefore, LO ⊥ MP

i.e., LN ⊥ MP, [Proved]

7. If the opposite sides of a quadrilateral are equal, prove that the quadrilateral will be parallelogram.

LMNO is a parallelogram quadrilateral, whose sides LM = ON and LO = MN. It is required to prove that LMNO is a parallelogram.

Construction: Diagonal LN is drawn.

Proof: In ∆LMN and ∆NOL,

LM = ON and MN = LO, [By hypothesis]

LN is common side.

Therefore, ∆LMN ≅ ∆NOL, [by Side Side Side congruence condition]

Therefore, ∠MLN = ∠LNO, [Corresponding angles of congruent triangles]

Since, LN cuts LM and ON and the both alternate angles are equal.

Therefore, LM ∥ ON

Again, ∠MNL = ∠OLN [Corresponding angles of congruent triangles]

But LN cuts LO and MN, and the alternate angles are equal.

Therefore, LO ∥ MN

LM ∥ ON and

LO ∥ MN.

Therefore, LMNO is a parallelogram. [Proved]

[Note: Rhombus is parallelogram.]

Congruent Line-segments

Congruent Angles

Congruent Triangles

Conditions for the Congruence of Triangles

Side Side Side Congruence

Side Angle Side Congruence

Angle Side Angle Congruence

Angle Angle Side Congruence

Right Angle Hypotenuse Side congruence

Pythagorean Theorem

Proof of Pythagorean Theorem

Converse of Pythagorean Theorem