Math Core

Right triangles come with a head start: every right angle is congruent to every other right angle, so one pair of angles is already matched before you look at anything else. That extra fact makes right triangles special in one surprising way. A side-side-angle combination, which fails for triangles in general, does work when the angle is a right angle.

Parts of a right triangle

In a right triangle, the side opposite the right angle is the hypotenuse. It is always the longest side. The two sides that form the right angle are the legs.

When you compare two right triangles, you get the pair of right angles for free. So right triangles need only two more pairs of parts, as long as they are the right pairs.

The hypotenuse-leg theorem

Right triangles with congruent hypotenuses (AB and DE) and congruent legs (AC and DF).

In the figure, both triangles are right triangles, the hypotenuses $\overline{AB}$ and $\overline{DE}$ are congruent, and the legs $\overline{AC}$ and $\overline{DF}$ are congruent. The right angle is not between the marked sides, so this looks like SSA. But here it works.

Hypotenuse-Leg (HL) theorem

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

Why it works. The Pythagorean theorem ties the three sides of a right triangle together: $a^2 + b^2 = c^2$ . If you know the hypotenuse $c$ and one leg $a$ , the other leg must be $b = \sqrt{c^2 - a^2}$ , with no other choice. So when two right triangles share a hypotenuse length and a leg length, their other legs are equal too, and the triangles are congruent by SSS.

For instance, any right triangle with hypotenuse $13$ and a leg of $5$ has other leg $\sqrt{169 - 25} = \sqrt{144} = 12$ . Every such triangle is a copy of the same $5$ - $12$ - $13$ triangle.

Using HL takes three things:

Both triangles are right triangles (say so in your proof).
The hypotenuses are congruent.
One pair of legs is congruent.

Common mistake

HL only works for right triangles. If neither angle is a right angle, two sides and a non-included angle do not prove congruence. Also, be sure the sides you match really are hypotenuse with hypotenuse and leg with leg. A hypotenuse congruent to a leg in the other triangle gives you nothing.

The other shortcuts still work

Everything you already know applies to right triangles, and the free right angle often completes a shortcut.

What you know (besides the right angles)	Shortcut
Hypotenuse and a leg	HL
Both legs	SAS (the right angle is included)
A leg and an acute angle	ASA or AAS
The hypotenuse and an acute angle	AAS

Some textbooks call these LL, LA and HA. They're not new postulates, just SAS, ASA and AAS with one pair of parts being the right angles.

Proofs with HL

Worked example: An altitude in an isosceles triangle

Given: $\overline{AD} \perp \overline{BC}$ and $\overline{AB} \cong \overline{AC}$

Prove: $\triangle ABD \cong \triangle ACD$

AD is perpendicular to BC, and AB = AC.

Statement	Reason
1. $\overline{AD} \perp \overline{BC}$	Given
2. $\angle ADB$ and $\angle ADC$ are right angles	Definition of perpendicular lines
3. $\triangle ABD$ and $\triangle ACD$ are right triangles	Definition of right triangle
4. $\overline{AB} \cong \overline{AC}$	Given
5. $\overline{AD} \cong \overline{AD}$	Reflexive property of congruence
6. $\triangle ABD \cong \triangle ACD$	HL

$\overline{AB}$ and $\overline{AC}$ are the hypotenuses (each is opposite a right angle at $D$ ), and $\overline{AD}$ is a leg of both triangles.

Worked example: A diagonal as the shared hypotenuse

Given: $\angle B$ and $\angle D$ are right angles, and $\overline{AB} \cong \overline{CD}$ .

Prove: $\triangle ABC \cong \triangle CDA$

Angles B and D are right angles, and AB = CD.

Statement	Reason
1. $\angle B$ and $\angle D$ are right angles	Given
2. $\triangle ABC$ and $\triangle CDA$ are right triangles	Definition of right triangle
3. $\overline{AC} \cong \overline{CA}$	Reflexive property of congruence
4. $\overline{AB} \cong \overline{CD}$	Given
5. $\triangle ABC \cong \triangle CDA$	HL

The diagonal $\overline{AC}$ is opposite both right angles, so it is the hypotenuse of each triangle. Notice the order in the statement: $\overline{AB}$ matches $\overline{CD}$ , so $A \leftrightarrow C$ and $B \leftrightarrow D$ , which leaves $C \leftrightarrow A$ .

Worked example: HL with algebra

$\triangle JKL$ and $\triangle MNP$ are right triangles with right angles at $K$ and $N$ . $JK = MN = 8$ , $JL = 17$ and $MP = 3x + 2$ . What value of $x$ makes the triangles congruent by HL, and what is the length of the other leg?

$\overline{JL}$ and $\overline{MP}$ are the hypotenuses (opposite the right angles). For HL they must be equal:

3x + 2 = 17 \quad\Rightarrow\quad x = 5.

The other leg of each triangle is $\sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15$ .

Tip

To spot the hypotenuse fast, find the right angle and look straight across the triangle from it.

Practice

Practice 1

$\triangle PQR$ and $\triangle XYZ$ are right triangles with right angles at $Q$ and $Y$ . $\overline{PR} \cong \overline{XZ}$ and $\overline{QR} \cong \overline{YZ}$ . Which shortcut proves $\triangle PQR \cong \triangle XYZ$ ?

Practice 2

$\triangle PQR$ and $\triangle XYZ$ are right triangles with right angles at $Q$ and $Y$ . $\overline{PQ} \cong \overline{XY}$ and $\overline{QR} \cong \overline{YZ}$ . Which shortcut proves $\triangle PQR \cong \triangle XYZ$ ?

Practice 3

$\triangle PQR$ and $\triangle XYZ$ are right triangles with right angles at $Q$ and $Y$ . $\overline{PR} \cong \overline{XZ}$ and $\angle P \cong \angle X$ . Which shortcut proves $\triangle PQR \cong \triangle XYZ$ ?

Practice 4

In $\triangle ABC$ and $\triangle DEF$ , $m\angle A = m\angle D = 70^\circ$ , $\overline{BC} \cong \overline{EF}$ and $\overline{AB} \cong \overline{DE}$ . Can you use HL to prove the triangles congruent?

Practice 5

In the proof that $\triangle ABD \cong \triangle ACD$ , where $\overline{AD} \perp \overline{BC}$ , what is the reason for the statement " $\angle ADB$ and $\angle ADC$ are right angles"?

Practice 6

$\triangle ABC \cong \triangle DEF$ by HL, with right angles at $C$ and $F$ . $AB = 25$ and $AC = 7$ . Find $EF$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

$\triangle JKL$ and $\triangle MNP$ are right triangles with right angles at $K$ and $N$ , and $\overline{JK} \cong \overline{MN}$ . If $JL = 4x - 1$ and $MP = 3x + 4$ , what value of $x$ makes the triangles congruent by HL?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

Two ladders, each $13$ feet long, lean against a vertical wall on level ground. The foot of each ladder is $5$ feet from the wall. Each ladder, the wall and the ground form a right triangle. Explain why the two triangles are congruent, then find how high up the wall each ladder reaches, in feet.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.