Side Side Side Postulate
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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


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.


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 = 90^{0}  12) Division property 
13) AP is the perpendicular bisector of BC  13) By definition of perpendicular bisector and from (9) and (12) 
• Side Angle Side Postulate
• Side Side Side Postulate
• Angle Angle Side Postulate
• Angle Side Angle Postulate
• HL postulate(Hypotenuse – Leg OR RHS)
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