Side Side Side Postulate

Side Side Side Postulate -> If the three sides of a triangle are congruent to the three sides of another triangle, then the two triangles are congruent.

AB ≅ DE , BC ≅ EF and AC ≅ DF
∴ ΔABC ≅ Δ DEF by SSS

Examples :

1) In triangle ABC, AD is median on BC and AB = AC.
Prove that ∠ABD = ∠ACD

Given : In ΔABC, AD is a median on BC and AB = AC.

Prove that : ∠ABD = ∠ACD

Statements	Reasons
1) AB = AC	1) Given
2) AD is a median	2) Given
3) BD = DC	3) By definition of median.
4) AD = AD	4) Reflexive (common side)
5) ΔADC ≅ ΔADB	5) By SSS postulate
6) ∠ABD = ∠ACD	6) CPCTC

2) ΔABC and ΔDBC are two isosceles triangle on the same base BC and vertices of A and D are on the same side of BC. If AD is extended to intersect BC at P.
Prove that i) ΔABD ≅ ΔACD ii) AP is the perpendicular bisector of BC.

Given : ΔABC and ΔDBC are two isosceles triangle.
⇒ AB = AC and BD = DC

Prove that : i) ΔABD ≅ ΔACD ii) AP is the perpendicular bisector of BC.

Statements	Reasons
1) AB = AC	1) Given
2) BD = CD	2) Given
3) AD = AD	3) Reflexive (common side)
4) ΔABD ≅ ΔACD	4) By SSS postulate
5) ∠BAP = ∠CAP	5) By CPCTC ∠BAD = ∠CAD
6) AP = AP	6) Reflexive(common side)
7) ΔBAP ≅ ΔCAP	7) By SAS postulate
8) ∠APB = ∠APC	8) CPCTC
9) BP = CP	9) CPCTC
10) ∠APB + ∠APC = 180	10) These two angles are linear pair angles and they are supplementary
11) 2∠APB = 180	11) Addition property
12) ang;APB = 900	12) Division property
13) AP is the perpendicular bisector of BC	13) By definition of perpendicular bisector and from (9) and (12)

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• Side Angle Side Postulate
• Side Side Side Postulate
• Angle Angle Side Postulate
• Angle Side Angle Postulate
• HL postulate(Hypotenuse – Leg OR RHS)

From side side side postulate to Postulates of Congruent Triangle

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