# Hypotenuse Leg Theorem – Explanation & Examples

In this article, we’ll learn about the **hypotenuse leg (HL) theorem**. Like, SAS, SSS, ASA, and AAS, it is also **one of the congruency postulates of a triangle.**

The difference is that the other 4 postulates apply to all triangles. Simultaneously, the **Hypotenuse Leg Theorem is true for the right triangles only** because, obviously, the hypotenuse is one of the right-angled triangle legs.

## What is Hypotenuse Leg Theorem?

The hypotenuse leg theorem is a criterion used to prove whether a given set of right triangles are congruent.

**The hypotenuse leg (HL) theorem states that; a given set of triangles are congruent if the corresponding lengths of their hypotenuse and one leg are equal. **

Unlike other congruency postulates such as; SSS, SAS, ASA, and AAS, three quantities are tested, with hypotenuse leg (HL) theorem, two sides of a right triangle are only considered.

*Illustration:*

### Proof of Hypotenuse Leg Theorem

In the diagram above, triangles *ABC* and* PQR* are right triangles with *AB* = *RQ*, *AC = PQ.*

By Pythagorean Theorem,

*AC ^{2} = AB^{2 }+ BC^{2}* and PQ

^{2}= RQ

^{2}+ RP

^{2}

Since *AC = PQ, *substitute to get;

*AB ^{2 }+ BC^{2} = RQ^{2} + RP^{2}*

But, *AB* = *RQ,*

By substitution;

*RQ ^{2} + *

*BC*

^{2}= RQ^{2}+ RP^{2}Collect like terms to get;

*BC ^{2} =RP^{2}*

Hence, *△**ABC **≅△** PQR*

**Example 1**

If *PR *⊥* QS, *prove that *PQR *and *PRS* are congruent

__Solution__

Triangle *PQR* and *PRS* are right triangles because they both have a 90-degree angle at point *R*.

Given;

*PQ = PS*(Hypotenuse)*PR = PR*(Common side)- Therefore, by Hypotenuse – Leg (HL) theorem,
*△**PQR**≅△**PR.*

**Example 2**

If *FB = DB,* *BA = BC*, *FB *⊥ *AE* and* DB* ⊥ *CE*, show that *AE = CE.*

__Solution__

By Hypotenuse Leg rule,

*BA = BC*(hypotenuse)*FB = DB*(equal side)- Since, ∆
*AFB*≅ ∆*BDC,*then ∠*A =**∠*Therefore,*AE = CE*

Hence proved.

*Example 3*

Given that ∆*ABC* is an isosceles triangle and ∠ *BAM = **∠**MAD*. Prove that *M *is the midpoint of *BD.*

__Solution__

Given ∠ *BAM = **∠**MAD*, then line AM is the bisector of ∠ *BAD.*

*AB = AD*(hypotenuse)*AM = AM*(common leg)- ∠
*AMB =**∠**AMD*(right angle) - Therefore,
*BM = MD.*

*Example 4*

Check whether ∆*XYZ* and ∆*STR *are congruent.

__Solution__

- Both ∆
*XYZ*and ∆*STR*are right triangles (presence of a 90 – degree angle) *XZ = TR*(equal hypotenuse).*XY = SR*(Equal leg)- Hence, by Hypotenuse-Leg (HL) theorem, ∆
*XYZ*≅∆*STR.*

*Example 5*

Given*: **∠**A=**∠**C = 90 *degrees*, AD= BC*. Show that △*ABD **≅* *△**DBC.*

__Solution__

Given,

*AD = BC*(equal leg)*∠**A=**∠**C*(right angle)*BD = DB*(common side, hypotenuse)- By, by Hypotenuse-Leg (HL) theorem, △
*ABD**≅**△**DBC*

*Example 6 *

Suppose ∠*W = **∠** Z* = 90 degrees and M is the midpoint of *WZ *and *XY. *Show that the two triangles *WMX* and* YMZ* are congruent.

__Solution__

- △
*WMX*and △*YMZ*are right triangles because they both have an angle of 90^{0 }(right angles) *WM = MZ*(leg)*XM = MY*(Hypotenuse)- Therefore
*,*by Hypotenuse-Leg (HL) theorem, △*WMX**≅*△*YMZ.*

*Example 7*

Calculate the value of x in the following congruent triangles.

__Solution__

Given the two triangles are congruent, then;

⇒2x + 2 = 5x – 19

⇒2x – 5x = -19 – 2

⇒ -3x = – 21

x =- 21/-3

x = 7.

Therefore, the value of x = 7

Proof:

⇒ 2x + 2 = 2(7) + 2

⇒14 + 2 = 16

⇒ 5x -19 = 5(7) – 19

⇒ 35 – 19 = 16

Yes, it worked!

*Example 8*

If *∠ **A = **∠ **C = 90 *degrees and *AD = BC.* Find the value of x and y that will make the two triangles *ABD *and *DBC *congruent.

__Solution__

Given,

△*ABD **≅* *△**DBC*

Calculate the value of x

⇒ 6x – 7 = 4x + 2

⇒ 6x – 4x = 2 + 7

⇒ 2x = 9

⇒ x = 9/2

x = 4.5

Calculate the value of y.

⇒ 4y + 25 = 7y – 5

⇒ 4y – 7y = – 5 – 25

⇒ -3y = -30

y = -30/-3 =10

Therefore, △*ABD **≅* *△**DBC*, when x = 4.5 and y = 2.72.