HL Postulate

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.

 Statements Reasons 1) AB = GE 1) Construction 2) ∠B = ∠FEG 2) Each 900 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 = 1800 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
 Statements Reasons 1) QM = MR 1) Given 2) LM = MN 2) Given 3)ML ⊥ PQ 3) Given 4) ∠QLM = 900 4) By definition of perpendicular 5) MN ⊥ PR 5) Given 6) ∠MNR = 900 6) By definition of perpendicular 7) ∠QLM = ∠MNR 7) Each 900 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)

From HL Postulate to Postulates of Congruent triangle