Side Angle Side Postulate
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.


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 .


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 congruent then 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.


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 90^{0} 
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)