What is the hypotenuse-leg congruence criterion?

Two right triangles are congruent if their hypotenuses are equal and one pair of legs is equal. It is also called HL, or RHS (right angle, hypotenuse, side).

How is hypotenuse-leg proved?

Lay the two triangles leg to leg on opposite sides of the equal leg. The two right angles make a straight line, the equal hypotenuses make an isosceles triangle with equal base angles, and then angle-angle-side applies. Alternatively, Pythagoras gives the other leg and SSS finishes.

Why does SSA work for right triangles but not in general?

With two sides and a non-included angle the third vertex can usually fall at either of two points. When the given angle is a right angle, the two points coincide in a single reflection, so the triangle is determined.

The Hypotenuse-Leg Criterion

How to use

▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
Slide $A$ along the line, or drag $B$ or $C$ ; the copy $A'$ follows, and the readout under the canvas keeps the angles at $A$ and $A'$ and the legs $CA$ and $CA'$ . Learn more about why it holds
Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene

Proof?Lay the second right triangle against the first along the equal legs, so that both share the leg BC and their other legs CA and CA′ point opposite ways. The two right angles at C make A, C, A′ a straight line, and the equal hypotenuses AB = A′B make ABA′ isosceles, so the angles at A and A′ are equal. Each triangle now has a right angle, the angle at its far vertex and the leg BC: angle-angle-side. Slide A along the line; drag B or C.HL congruence

DisplayLabelsMeasures

Slide A along the line; drag B or C

∠A = 55.8°,  ∠A′ = 55.8°,  CA = 2.3,  CA′ = 2.3

Step 1 / 5

Step explanations

1Two right triangles, leg to leg

Given
ABC and A′BC are right-angled at C, with the hypotenuses AB = A′B (one tick each) and the leg BC in common (two ticks). A′BC is the second triangle of the theorem, moved so that its leg equal to BC lies on BC and its far vertex A′ falls on the other side. To prove: ABC ≅ A′BC.

A′ is the mirror of A through C; slide A and it follows. Learn more about placing the triangles

Getting Started with the Visualizer

The Statement

Leg to Leg

A Straight Line

The Isosceles Triangle

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 6

Step through the proof, then slide

A

along the line and drag

B

C

, and watch the copy keep the same angle at its far vertex and the same other leg. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.

Opening scene: the controls

Proof?HL congruenceDisplayLabelsMeasuresSlide A along the line; drag B or C

∠A = 55.8°,  ∠A′ = 55.8°,  CA = 2.3,  CA′ = 2.3

← Back▶ PlayNext →ResetNormal ▾Step 1 / 5

Step explanations

1Two right triangles, leg to leg

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1Proof — the theorem loaded in the player; the ? gives its idea in one line

2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc

3Draggable points — B and C move freely, A slides along the perpendicular at C; the hollow vertex A′ is the mirror of A through C

4Readout — the line under the figure: the angles at A and A′ and the legs CA and CA′, live while you drag

5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure

6Speed and step counter — how fast Play advances, and which of the 5 steps is on screen

7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it

Proof complete: step 5 of 5

Proof?HL congruenceDisplayLabelsMeasuresSlide A along the line; drag B or C

∠A = 55.8°,  ∠A′ = 55.8°,  CA = 2.3,  CA′ = 2.3

← Back↻ ReplayNext →ResetNormal ▾Step 5 / 5

Step explanations

1Two right triangles, leg to leg

2A, C and A′ lie on one line

Straight angle
The two right angles at C, ∠BCA and ∠BCA′, are on opposite sides of BC and add up to 180°, a straight angle. So CA and CA′ are opposite rays, and A, C, A′ lie on one line (dashed).

3∠BAA′ = ∠BA′A = α

Isosceles base angles
Look at the large triangle ABA′. Its sides AB and A′B are the two equal hypotenuses, so it is isosceles with apex B, and its base angles at A and at A′ are equal. Since C lies on the base AA′, these are the angles ∠BAC of the first triangle and ∠BA′C of the second.

4ABC ≅ A′BC

AAS congruence
In ABC and A′BC: the angles at A and A′ are equal (α), the angles at C are both right angles, and the side BC, opposite α in both, is common. Two angles and a side not between them: the triangles are congruent by angle-angle-side.

5Hypotenuse and leg determine a right triangle

Conclusion
A′BC is a copy of the second given triangle, so the two given triangles are congruent: in right triangles, an equal hypotenuse and one equal leg are enough. In particular the other legs agree, CA = CA′. ∎

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1The common leg — BC, two ticks: the equal legs laid on each other, with a right angle on each side at C

2The hypotenuses — AB and A′B, one tick each: the equal sides of the isosceles triangle ABA′

3The base angles — α at A and at A′: equal, and the angles AAS needs

4Justification tags — Given, Straight angle, Isosceles base angles, AAS congruence, Conclusion

5Replay — at the last step Play becomes Replay and Next is off

The Statement

Two sides and an angle not between them do not determine a triangle in general; the AAS congruence page explains why. Right triangles are the exception. If the angle is the right angle, the two sides are the hypotenuse and one leg, and they are enough:

\angle C = \angle C' = 90°, \; AB = A'B', \; BC = B'C' \implies ABC \cong A'B'C'

This is the hypotenuse-leg criterion, also written HL or, in some countries, RHS (right angle, hypotenuse, side). The proof here lays the second triangle against the first along the equal legs, so that the two hypotenuses become the equal sides of one isosceles triangle.

In the player the right angle is at

C

BC

is the common leg,

AB

and

A'B

are the hypotenuses, and

A'

is placed as the mirror image of

A

through

C

. The readout under the canvas shows the angles at

A

and

A'

and the two other legs

CA

and

CA'

Leg to Leg

Take the equal legs first. The second triangle has a leg

B'C'

equal to

BC

; move it so that this leg lies on

BC

, with

C'

C

and

B'

B

, and so that its far vertex falls on the other side of the line

BC

from

A

. Call that vertex

A'

. Then

A'BC

is right-angled at

C

and

A'B = A'B' = AB

, marked in the player with one tick on each hypotenuse and two ticks on the common leg.

The two triangles now sit leg to leg, one on each side of

BC

, filled in two colours. The claim to prove is that they are congruent, and in particular that

CA' = CA

Opening scene, frozen

ABC and its copy A′BC leg to leg: right angles at C, AB = A′B, BC in common.

A Straight Line

C

there are two right angles,

\angle BCA

on one side of

BC

and

\angle BCA'

on the other. Together they make

\angle BCA + \angle BCA' = 90° + 90° = 180°

a straight angle, so

CA

and

CA'

are opposite rays and

A

C

A'

lie on one line. The player draws it dashed. This is the same fact about a straight angle that the vertical angles page uses, read the other way round.

A straight line, frozen

Two right angles at C make a straight angle: A, C, A′ are collinear.

The Isosceles Triangle

Now look at the large triangle

ABA'

made by the two hypotenuses and the line through

C

. Its sides

AB

and

A'B

are the two hypotenuses, which are equal, so

ABA'

is isosceles with apex

B

, and by the isosceles base angles theorem its base angles are equal:

\angle BAA' = \angle BA'A = \alpha

Because

C

lies on the base

AA'

, these two angles are exactly the angles of the two right triangles at their far vertices:

\angle BAC = \alpha

and

\angle BA'C = \alpha

The isosceles triangle, frozen

ABA′ has two equal hypotenuses as sides, so the base angles at A and A′ are both α.

Angle-Angle-Side

Compare the two right triangles once more:

\angle A = \angle A' = \alpha, \qquad \angle C = \angle C' = 90°, \qquad BC = BC

Two angles agree, and the side

BC

, which is opposite

\alpha

in both triangles, is common. That is the AAS congruence pattern, so

ABC \cong A'BC

Angle-angle-side, frozen

α, the right angle and the common leg BC: ABC ≅ A′BC by AAS.

Why It Holds

Since

A'BC

is the second given triangle moved into place, the two given triangles are congruent: in right triangles the hypotenuse and one leg determine everything else, including the other leg,

CA = CA'

. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given data, a straight angle, the isosceles theorem, and angle-angle-side.

There is a second proof, by computation. The Pythagorean theorem gives the other leg from the hypotenuse and the known leg,

CA^2 = AB^2 - BC^2

, and the same in the second triangle, so

CA = C'A'

; then all three sides agree and SSS congruence finishes. The proof shown here is older and uses no lengths at all.

The conclusion, frozen

Hypotenuse and leg determine a right triangle; in particular CA = CA′.

Dragging the Figure

Slide

A

along the line through

C

: the first triangle gets a longer or shorter leg

CA

, and the copy follows with

CA' = CA

, since

A'

is always the mirror of

A

through

C

; the readout keeps the angles at

A

and

A'

equal. Drag

B

to change the common leg and the hypotenuses, or

C

to move the right angle;

A

stays on the perpendicular to

BC

C

.

A drag that would flatten either triangle, shorten the common leg too far, or carry a vertex out of the frame is refused.

After dragging A and B, frozen

A slid towards C and B raised: shorter legs, steeper hypotenuses, the angles at A and A′ still equal.

Consequences

• Used on these pages. Hypotenuse-leg is the usual way to show the two tangents from a point are equal (the radii are legs, the segment to the centre a common hypotenuse) and that the perpendicular from the centre bisects a chord (two radii as hypotenuses, the perpendicular a common leg). Those pages used Pythagoras instead; either route works. It also gives the equal tangent lengths on the incentre page.
• The SSA exception. Two sides and a non-included angle fail in general because the third vertex can fall at two places. When the angle is right, one of the two places disappears, and the triangle is determined.
• Five criteria. Side-angle-side, side-side-side, angle-side-angle, angle-angle-side and hypotenuse-leg: every congruence criterion used in school geometry now has a page here, all derived from side-angle-side.
• Perpendicular bisector, again. A point equidistant from two points lies on their perpendicular bisector: drop the perpendicular to the segment, and hypotenuse-leg makes the two halves equal.
• Reflections. The proof places

A'

as the mirror image of

A

C

along the line; the theorem says that any right triangle with the same hypotenuse and leg is that mirror image, up to moving it rigidly.