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
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| 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.
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| 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)
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