To Prove Hypotenuse Leg theorem we need to construct an additional triangle.

**Given:** ΔABC and ΔDEF are right triangles with right angles at B and E.

AB ≅

DE and

AC ≅

DF**Prove:** ΔABC ≅ ΔDEF.

**Proof:** On ΔDEF extend the side FE to FG such that BC = EG.

Considering Δs ABC and DEG,

AB ≅

DE (Given)

BC ≅

EG (By construction)

∠B ≅ ∠E (Right angles)

ΔABC ≅ Δ DEG (By SAS criterion for congruency).

AC ≅

DG (CPCTC)

DG ≅

DF (Transitive Property of congruence and given

AC ≅

DF).

Hence ΔGDF is an isosceles triangle.

Considering Δs DEG and DEF,

∠G ≅ ∠F (By Isosceles triangle theorem)

∠DEG ≅ ∠DEF (Right angles)

DE ≅ DE (Reflexive property of congruence)

ΔDEG ≅ ΔDEF (By AAS criterion of congruency)

Hence

ΔABC ≅ ΔDEF (Transitive property of congruence)