Hypotenuse Leg theorem states a condition for two right triangles to be congruent. Generally to check for triangle congruence three corresponding parts are considered. In the case of right triangles, as one corresponding part the right angle is implied in the name, the names of congruence criteria consist of only two parts as hypotenuse leg in this case. The theorem is also known shortly as HL Theorem.
let us state, prove the hypotenuse leg theorem and solve few problems based on the theorem.

## Hypotenuse Leg Congruence Theorem

If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle, then the two triangles are congruent.

If Δs ABC and DEF are right triangles and leg AB ≅ leg DE and hypotenuse AC = hypotenuse DF,
then ΔABC ≅ ΔDEF.

## Prove Hypotenuse Leg theorem

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.
ABDE and ACDF
Prove: ΔABC ≅ ΔDEF.
Proof: On ΔDEF extend the side FE to FG such that BC = EG.
Considering Δs ABC and DEG,
ABDE (Given)
BC EG (By construction)
∠B ≅ ∠E (Right angles)
ΔABC ≅ Δ DEG (By SAS criterion for congruency).
ACDG (CPCTC)
DGDF (Transitive Property of congruence and given ACDF).
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)

## Hypotenuse Leg Theorem Proof 2

Hypotenuse Leg theorem can also be proved using Pythagorean theorem. According to Pythagorean theorem if ΔABC is a right triangle right angled at the vertex C, then c2 = a2 + b2, where 'a' and 'b' are the lengths of the legs and 'c' the length of the hypotenuse. Using this relationship, the length of one of the legs can be solved as a2 = c2 - b2.

Given: ΔABC and ΔDEF are right triangles with right angles at B and E.
ABDE and ACDF
Prove: ΔABC ≅ ΔDEF.
Proof: b and f are the lengths of hypotenuses as shown in the figure. Hence we have the relationships using Pythagorean
theorem as,
For ΔABC, b2 = a2 + c2.
For ΔDEF, e2 = d2 + f2.
Solving for the leg we have,
a2 = b2 - c2 and d2 = e2 - f2.
a2 = b2 - c2 = e2 - f2 = d2 (by substitution Lengths of congruent sides)
Hence a = d ⇒ BCEF
Hence ΔABC ≅ ΔDEF (by SSS criterion of congruency)
Indeed it can be seen that the Hypotenuse Leg congruence theorem is equivalent of SSS congruence theorem, as the congruence of the third corresponding sides is implied when it is given two corresponding sides are congruent.

## Hypotenuse Leg Theorem Examples

1. Which of the given information can be used to prove ΔAOB ≅ ΔCOD using Hypotenuse Leg Theorem

1. m ∠AOD = 90º
2. AB > OB
3. AODO
4. AO CO

It is shown in the diagram, the two hypotenuse are congruent in the two right triangles. Hence we need to look for a statement that tells of leg congruence. There are two options that state the leg congruence, statements 3 and 4. The congruence statement given in statement 3 is not for corresponding legs when the names of the triangles are considered as AOD and COD. Hence the information AOCO is the additional information needed to prove
ΔAOB ≅ ΔCOD using HL theorem.

2. Prove in an isosceles triangle, the altitude from the vertex divides the triangle into two congruent triangles.

Given: ΔABC is isosceles such that ABAC and AD the altitude from A.
Prove: ΔABD ≅ ΔACD

 Statement Reason 1. AD ⊥ BC  2. ∠ADB and ∠ADC are right angles  3. AB ≅ AC  4. AD ≅ AD  5. ΔABD ≅ ΔACD 1. Definition of altitude2. Definition of perpendicular3. Given4. Reflexive property of congruence5. HL theorem of congruence.

3. In the diagram given below, ABDE, CEFB and m ∠CAB = m ∠FDE = 90º.
Write a paragraph proof to prove ΔCAB ≅ ΔFDE.

From the figure it can be seen, two corresponding legs are congruent for the two right triangles to be considered.
Length of Hypotenuse BC = CE + BE by segment addition postulate. Substituting FB for CE we get,
Length of Hypotenuse BC = FB + BE = Hypotenuse EF. Hence BCEF. Hence by hypotenuse leg theorem

ΔABC ≅ Δ DEF.