Hypotenuse Leg Theorem – Definition, Proof, Examples, FAQs

What Is the Hypotenuse Leg Theorem?

Hypotenuse Leg Theorem (HL theorem) is used to prove the congruence of two right angled triangles. It is also known as the RHS (Right angle-Hypotenuse-Side) congruence rule. 

In a right triangle, there is one right angle ($90^{\circ}$ angle) and two acute angles. The side opposite to the right angle is called hypotenuse. It is the longest side of the right triangle. The other two sides are called ‘legs’ of the right triangle.

According to the HL theorem, if the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of the other right triangle, then the two triangles are congruent.

Hypotenuse Leg Theorem Statement

HL triangle congruence theorem: If the hypotenuse and one leg of one right triangle are congruent to the hypotenuse and leg of the other right triangle, then the two right triangles are said to be congruent. 

In other words, two right triangles are congruent if the corresponding lengths of the hypotenuse and one leg are equal.  

Hypotenuse Leg Theorem Proof

Given: In right triangles △ABC ≅△XYZ, AB = XY and AC = XZ

To prove: △ABC ≅△XYZ

Proof: 

Statement	Reason
AB2 + BC2 = AC2	Pythagoras theorem
XY2 + YZ2 = XZ2	Pythagoras theorem
XZ = AC                                            …(1)	Given
AB2 + BC2 = XY2 + YZ2                                    …(2)	From (1)
AB = XY                                            …(3)	Given
XY2 + BC2 = XY2 + YZ2	From (2) and (3)
BC2 = YZ2	Canceling out XY2
BC = XY	Taking positive square root
△ABC ≅△XYZ	By SSS criterion

Facts about Hypotenuse Leg Theorem

Conclusion

In this article, we learned about HL theorem or HL congruence criterion to prove congruence of two right triangles. We learned the statement and also proved the theorem. Let’s use the HL theorem to solve a few examples and practice problems.

Solved Examples on HL Theorem (Hypotenuse Leg Theorem)

1. Is △PQR ≅ △NLM? 

Solution: 

To prove the congruence of right triangles, we will use the HL theorem.

In right triangles △PQR and △LMN, we have

∠Q = ∠L = 90°

PR = MN = 13 units

QR = LM = 5 units

By HL Theorem, one leg and hypotenuse of △PQR are congruent to one leg and hypotenuse of △LMN.

Thus, △PQR ≅ △NLM

2. In the △PQR, if PS is the perpendicular bisector of the side QR, then prove that ∠QPS = ∠RPS.

Solution: 

In △PQR, PS is the perpendicular bisector of QR.

$\Rightarrow$ QS = SR …(1)

In △PSQ and △PSR

∠PSQ = ∠PSR = 90°

PQ = PR = 3 units

QS = SR …from (1)

Thus, △PSQ ≅ △PSR …by HL Theorem

$\Rightarrow$ ∠QPS = ∠RPS …CPCTC (Corresponding parts of congruent triangles are congruent.)

3. In the following figure, find the value of JK and MN.

Solution: 

In △IJK and △ONM,

∠IJK = ∠ONM = 90°

IK = MO = 5 units

IJ = ON = 4 units

By HL Theorem,

△IJK ≅ △ONM

$\Rightarrow$KJ = MN (CPCTC)

KJ = MN = $\sqrt{5^{2}\;-\;4^{2}}$ (By Pythagoras theorem)

KJ = MN = $\sqrt{25 \;-\; 16}

KJ = MN = $\sqrt{9}$

KJ = MN = 3 units

4. In the following figure, prove that △DAB ≅ △CBA.

Solution: 

In △DAB and △CBA, we have

∠ADB = ∠ACB = 90°

AB = AB (common side)

AD = BC = 4 units

By HL Theorem,

△DAB ≅ △CBA

5. In the following figure, prove that △ABC ≅ △EDC.

Solution: 

CD = BD – BC = 14 – 8 = 6 units

In △ABC, we have

$AC^{2} = AB^{2} + BC^{2}$

$AC^{2} = 6^{2} + 8^{2} = 36 + 64 = 100$

AC = 10 inches

In △ABC and △EDC,

∠ABC = ∠EDC = 90°

AC = CE = 10 units …hypotenuse

AB = CD = 6 units …leg

By HL Theorem,

△ABC ≅ △EDC

Practice Problems on HL Theorem (Hypotenuse Leg Theorem)

Frequently Asked Questions on HL Theorem (Hypotenuse Leg Theorem)