# 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 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)

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