side angle side postulate

# Side Angle Side Postulate

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Side angle side postulate -> If two sides and the included angle of a triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent by side angle side postulate. AB ≅ DE, BC ≅ EF and ∠B ≅ ∠E
∴ ΔABC ≅ Δ DEF by SAS

Theorem : Angles opposite to two equal sides of a triangle are equal.

Given : Δ ABC in which AB = AC

Prove that : ∠C = ∠B Construction : Draw the bisector AD of ∠A which meets BC in D.

 Statements Reasons 1) AB = AC 1) Given 2) AD is a bisector 2) By construction 3) ∠BAD = ∠CAD 3) By definition of angle bisector 4) AD = AD 4) Reflexive (common side) 5) ΔABD ≅ ΔACD 5) SAS Postulate 6) ∠B = ∠C 6) CPCTC

Examples :

1) O is the mid point of AB and CD. Prove that
i) ΔAOC ≅ ΔBOD ii) AC = BD and iii) AC || BD .

Given : O is the mid point of AB and CD.

Prove that : i) ΔAOC ≅ ΔBOD ii) AC = BD and iii) AC || BD . Statements Reasons 1) O is the mid point. 1) Given 2) AO = OB 2) By definition of mid point. 3) ∠AOC = ∠BOD 3) Vertically opposite angles 4) CO = OD 4) By definition of mid point. 5) ΔAOC ≅ ΔBOD 5) SAS postulate 6) AC = BD 6) CPCTC 7) ∠CAO = ∠DBO 7) CPCTC 8) AC || BD 8) If alternate interior angles are congruentthen the lines are parallel.

2) If D is the mid point of the hypotenuse AC of a right triangle ABC, prove that BD = ½ AC.

Given : ΔABC in which ∠B = 90 0 and D is the mid point of AC.

Prove that : BD = ½ AC.

Construction : Produce BD to E so that BD = DE. Join EC. Statements Reasons 1) AD = DC 1) Given 2) BD = DE 2) By construction 3) ∠ADB = ∠CDE 3) Vertically opposite angles 4) ΔADB ≅ ΔCDE 4) By SAS postulate 5) EC = AB and ∠CED = ∠ABD 5) CPCTC 6) CE || AB 6) If alternate interior angles are congruent then the lines are parallel 7) ∠ABC + ∠ECB = 180 7) Angles formed on the same side of transveral are supplementary. 8) 90 + ∠ECB = 180 8) Since ∠B = 90 given 9) ∠ECB = 180 -90 = 90 9) By subtraction property 10) AB = EC 10) From (5) 11) BC = CB 11) Reflexive (Common side) 12) ∠ABC = ∠ECB 12) Each 900 13) ΔABC ≅ ΔECB 13) SAS postulate 14) AC = BE 14) CPCTC 15) 1/2AC = 1/2BE ⇒ 1/2AC = BD 15) Multiply by 1/2 but 1/2BE = BD by mid point definition.

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

From side angle side postulate to Postulates of Congruent Triangle

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