HL Postulate
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HL Postulate(Hypotenuse  Leg) or RHS theorem > If any two right angles that have a congruent hypotenuse and a corresponding, congruent leg are congruent triangles.AC ≅ DF , BC ≅ EF and ∠B ≅ ∠E ( both 90 ^{0} )
∴ Δ ABC ≅ Δ DEF by RHS theorem
Theorem : Prove RHS theorem or postulate.
Given : AC = DF , BC = EF and ∠B = ∠E = 90 ^{0}
Prove that : ΔABC ≅ ΔDEF
Construction : Extend DE to G so that EG = AB. Join GF.


1) AB = GE  1) Construction 
2) ∠B = ∠FEG  2) Each 90^{0} 
3) BC = EF  3) Given 
4) ΔABC ≅ ΔGEF  4) By SAS postulate 
5) ∠A = ∠G  5) CPCTC 
6) AC = GF  6) If the two angles are congruent then angle opposite to them are equal 
7) AC = DF  7) Given 
8) DF = GF  8) By transitive property 
9) ∠D = ∠G  9) Angles opposite to equal sides in ΔDGF are equal 
10) ∠A = ∠D  10) From (5) and (9) 
11) ∠B = ∠E  11) Given 
12)∠A+∠B=∠D+∠E  12) Adding (10) and (11) 
13) ∠C = ∠F 
13) ∠A + ∠B + ∠C = 180 and ∠D + ∠E + ∠F = 180^{0} 
14) ΔABC ≅ ΔDEF  14) By SAS postulate and from (3) (7) and (13) 
1)
Given : LM = MN , QM = MR, MR ⊥ PQ and MN ⊥ PR.
Prove that : PQ = PR


1) QM = MR  1) Given 
2) LM = MN  2) Given 
3)ML ⊥ PQ  3) Given 
4) ∠QLM = 90^{0}  4) By definition of perpendicular 
5) MN ⊥ PR  5) Given 
6) ∠MNR = 90^{0}  6) By definition of perpendicular 
7) ∠QLM = ∠MNR  7) Each 90^{0} 
8) ΔQLM ≅ ΔRNM  8) RHS theorem 
9) ∠Q = ∠R  9) CPCTC 
10) PR = PQ  10) If two angles are equal then side opposite to them are also equal 
• 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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