Geometry · G.CO.B.8 · G.SRT.B.5

Hypotenuse leg congruence

Side, side and an angle not between them usually fail to fix a triangle. When the angle is a right angle, do they?

You need

The Pythagorean theoremMath 8 SSS, SAS, ASA, AASGeometry

Where you are 3 in Congruent triangles

Geometry11

Triangles18

Congruent triangles3

Congruence statements and correspondence Why AAA and SSA fail

Hypotenuse leg congruence

Why this exists

SSS, SAS, ASA and AAS each name three matching parts that force two triangles to be congruent. SSA is not on the list: two sides and an angle that is not between them can fit two different triangles (why side side angle fails).

Right triangles are the exception that matters most. The right angle often comes for free, and the lengths easiest to measure are the longest side and one other: a wire and the height of the pole it holds. That is side, side and an angle not between them, and for right triangles it is enough, because the Pythagorean theorem makes two sides of a right triangle decide the third. This lesson proves it as the HL theorem.

Definition

In a right triangle the two sides that meet at the right angle are the legs, and the side opposite the right angle is the hypotenuse.

Hypotenuse leg congruence, or HL, says: two right triangles are congruent when the hypotenuse of one is congruent to the hypotenuse of the other and a leg of one is congruent to a leg of the other. All three facts, the right angles included, must be known, not just seen in a drawing.

Read around the triangle, those are an angle (the right angle), a side touching it (the leg) and a side facing it (the hypotenuse). The angle is not between the two sides, so HL is the right-angle case of SSA, and that case is safe (it is not the only one: see Boundaries). SSS, SAS and ASA are postulates on this map; HL is a theorem, proved below from the Pythagorean theorem and the SSS postulate.

Intuition

Build a right triangle from a leg and a hypotenuse. Stand the leg \overline{CB} straight up from a level line, so that the right angle is at C. The third vertex has to be on the line. The hypotenuse starts at B with a known length and an unknown direction, like a stick hinged at B. Swing it down until its free end touches the line.

Figure 1The hypotenuse, hinged at B, lands at A or at A'. The two triangles are mirror images of each other.

It lands twice, at A and at A' (Figure 1). In the SSA picture the two landings sat on the same ray, either side of the foot of the perpendicular from B, and made a narrow triangle and a wide one. Here that foot is C itself, so the landings sit equally far from C on opposite sides: \triangle ABC and \triangle A'BC are one triangle flipped over.

Results

HL 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 triangles are congruent.

In Figure 2, \angle C and \angle F are right angles, \overline{AB} \cong \overline{DE} and \overline{BC} \cong \overline{EF}, so

\triangle ABC \cong \triangle DEF \quad\text{by HL.}

Figure 2What HL asks for. A right angle in each, the hypotenuses, and one pair of legs.

The unmarked legs then match, \overline{AC} \cong \overline{DF}, and so do the acute angles, because corresponding parts of congruent triangles are congruent (its own lesson).

Some books add LL (two legs), LA (a leg and an acute angle) and HA (the hypotenuse and an acute angle). None is new: with the right angles congruent by the Right Angle Congruence Theorem, LL is SAS, HA is AAS, and LA is ASA or AAS as the leg does or does not touch the acute angle.

Why it is true

Given. \angle C and \angle F are right angles; \overline{AB} \cong \overline{DE}; \overline{BC} \cong \overline{EF} (Figure 2).

Prove. \triangle ABC \cong \triangle DEF.

Plan. Two pairs of sides are given, so SSS needs only the third pair, \overline{AC} and \overline{DF}. Write the Pythagorean theorem in each triangle, swap in the given equal lengths, and the squares of the third sides come out equal.

Statements	Reasons
1. \angle C and \angle F are right angles	given
2. \overline{AB} \cong \overline{DE}	given
3. \overline{BC} \cong \overline{EF}	given
4. AB = DE	definition of congruent segments: line 2
5. BC = EF	definition of congruent segments: line 3
6. AB^2 = AC^2 + BC^2	Pythagorean theorem: \triangle ABC, line 1
7. DE^2 = DF^2 + EF^2	Pythagorean theorem: \triangle DEF, line 1
8. DE^2 = AC^2 + EF^2	substitution: lines 4 and 5 into line 6
9. AC^2 + EF^2 = DF^2 + EF^2	substitution: line 8 into line 7
10. AC^2 = DF^2	subtraction property of equality: line 9
11. AC = DF	square root property: line 10; AC and DF are lengths, so positive
12. \overline{AC} \cong \overline{DF}	definition of congruent segments: line 11
13. \triangle ABC \cong \triangle DEF	SSS postulate: lines 2, 3 and 12

The square root property allows AC = DF or AC = -DF; line 11 keeps the first because no length is negative. Lines 8 to 11 are algebra: the geometry is all in the Pythagorean theorem, which is why HL needs right triangles. That theorem is proved on this map by rearranging areas, with no congruence criterion at all, so nothing here is circular.

Worked examples

Example 1

HL or not

\triangle PQR and \triangle XYZ are right triangles with right angles at R and Z. For each pair of facts, say whether HL proves the triangles congruent. If it does, write the congruence statement; if not, say what does, or that nothing does.

(a) \overline{PQ} \cong \overline{XY} and \overline{QR} \cong \overline{YZ}

(b) \overline{PQ} \cong \overline{XY} and \overline{PR} \cong \overline{YZ}

(d) \overline{PQ} \cong \overline{YZ} and \overline{QR} \cong \overline{XZ}

Solution

First find the hypotenuses, the sides opposite the right angles: \overline{PQ} faces \angle R and \overline{XY} faces \angle Z. The legs are \overline{PR}, \overline{QR}, \overline{XZ} and \overline{YZ}.

(a) Hypotenuse with hypotenuse, leg with leg: HL. The leg \overline{QR} meets the hypotenuse at Q, and \overline{YZ} meets it at Y, so Q \leftrightarrow Y, R \leftrightarrow Z and P \leftrightarrow X: \triangle PQR \cong \triangle XYZ.

(b) HL again, but the given leg changes the pairing. \overline{PR} meets the hypotenuse at P and \overline{YZ} meets it at Y, so P \leftrightarrow Y, R \leftrightarrow Z and Q \leftrightarrow X:

\triangle PQR \cong \triangle YXZ

Writing \triangle PQR \cong \triangle XYZ would claim \overline{PR} \cong \overline{XZ}, which nobody gave.

(c) Two legs, no hypotenuse: not HL. The right angles, congruent to each other, are included between the legs, so SAS gives \triangle PQR \cong \triangle XYZ.

(d) \overline{PQ} is a hypotenuse and \overline{YZ} is a leg, so nothing applies. In fact the triangles cannot be congruent: the hypotenuse is the longest side of a right triangle (its square is the sum of the other two squares), so XY > YZ = PQ, and congruent triangles have the same longest side.

Example 2

Solving for the values that make HL work

\triangle JKL and \triangle MNP are right triangles with right angles at L and P. Also JK = 3x + 2, MN = 5x - 8, KL = 2y and NP = y + 4. Find x and y so that HL proves \triangle JKL \cong \triangle MNP, then find JL.

Solution

\overline{JK} faces \angle L and \overline{MN} faces \angle P, so they are the hypotenuses; \overline{KL} and \overline{NP} are legs. HL needs each pair equal:

\begin{aligned} 3x + 2 &= 5x - 8 \\ 10 &= 2x \\ x &= 5 \end{aligned}

2y = y + 4 \quad\Longrightarrow\quad y = 4

So JK = MN = 17 and KL = NP = 8. The Pythagorean theorem in \triangle JKL gives the third side:

\begin{aligned} JL^2 + 8^2 &= 17^2 \\ JL^2 &= 289 - 64 = 225 \\ JL &= 15 \end{aligned}

Check. 3(5) + 2 = 17 = 5(5) - 8, 2(4) = 8 = 4 + 4, and 15^2 + 8^2 = 225 + 64 = 289 = 17^2. By corresponding parts, MP = 15 as well.

Example 3

A pole and two wires

A pole stands straight up on level ground. Two wires, each 6.5\text{ m} long, run from the point P, 6\text{ m} up the pole, to stakes A and B on opposite sides (Figure 3). Show that the stakes are the same distance from the foot F of the pole and that the two wires meet the ground at the same angle. How far is each stake from F?

Figure 3Two equal wires from one point on a pole.

Solution

The pole meets the level ground at right angles, so \angle PFA and \angle PFB are right angles. The hypotenuses \overline{PA} and \overline{PB} are both 6.5 m, and the leg \overline{PF} belongs to both triangles, so it is congruent to itself. HL gives

\triangle PFA \cong \triangle PFB

with P \leftrightarrow P, F \leftrightarrow F, A \leftrightarrow B. Corresponding parts of congruent triangles are congruent, so \overline{FA} \cong \overline{FB} and \angle PAF \cong \angle PBF. For the distance, use the Pythagorean theorem in \triangle PFA:

\begin{aligned} FA^2 + 6^2 &= 6.5^2 \\ FA^2 &= 42.25 - 36 = 6.25 \\ FA &= 2.5 \end{aligned}

Each stake is 2.5 m from the foot of the pole.

Example 4

The altitude of an isosceles triangle

Given. \overline{AB} \cong \overline{AC}; \overline{AD} \perp \overline{BC}, with D on \overline{BC} (Figure 4).

Prove. \overline{BD} \cong \overline{CD}.

Figure 4\overline{AB} \cong \overline{AC} and \overline{AD} \perp \overline{BC}. Where does D land?

Solution

Plan. \overline{AD} cuts the triangle into \triangle ABD and \triangle ACD, right-angled at D. Their hypotenuses \overline{AB} and \overline{AC} are given congruent, and they share the leg \overline{AD}: HL. The Prove is then a pair of corresponding parts.

Statements	Reasons
1. \overline{AB} \cong \overline{AC}	given
2. \overline{AD} \perp \overline{BC}	given
3. \angle ADB and \angle ADC are right angles	four right angles: line 2
4. \overline{AD} \cong \overline{AD}	reflexive property of congruence
5. \triangle ABD \cong \triangle ACD	HL theorem: lines 3, 1 and 4
6. \overline{BD} \cong \overline{CD}	corresponding parts of congruent triangles: line 5

So D is the midpoint of \overline{BC}, which the Pythagorean theorem lesson took as given. Proving the Base Angles Theorem reaches the same midpoint from the altitude only after that theorem, by AAS (its Example 3); HL needs no angle but the right one, and line 5 hands over \angle B \cong \angle C as well.

Example 5

A point the same distance from both sides of an angle

Given. P is inside \angle XOY; \overline{PA} \perp \overrightarrow{OX} at A; \overline{PB} \perp \overrightarrow{OY} at B; \overline{PA} \cong \overline{PB} (Figure 5).

Prove. \overrightarrow{OP} bisects \angle XOY.

Figure 5P is the same distance from both sides of the angle.

Solution

Plan. The distance from a point to a line is measured along the perpendicular (distance from a point to a line), so PA and PB are the distances from P to the two sides, and the right angles at A and B come with them. The right triangles \triangle OAP and \triangle OBP share the hypotenuse \overline{OP} and have congruent legs.

Statements	Reasons
1. \overline{PA} \perp \overrightarrow{OX}	given
2. \overline{PB} \perp \overrightarrow{OY}	given
3. \angle OAP is a right angle	four right angles: line 1
4. \angle OBP is a right angle	four right angles: line 2
5. \overline{PA} \cong \overline{PB}	given
6. \overline{OP} \cong \overline{OP}	reflexive property of congruence
7. \triangle OAP \cong \triangle OBP	HL theorem: lines 3, 4, 6 and 5
8. \angle AOP \cong \angle BOP	corresponding parts of congruent triangles: line 7
9. \overrightarrow{OP} bisects \angle XOY	definition of angle bisector: line 8, P inside \angle XOY

Line 8 is \angle XOP \cong \angle YOP, since A lies on \overrightarrow{OX} and B on \overrightarrow{OY}. So a point inside an angle, equally far from its two sides with the perpendiculars landing on the sides as here, lies on the bisector.

Boundaries

The right angles must be known: given, marked, or proved on an earlier line. Without them the data is ordinary SSA and the Pythagorean theorem has nothing to act on.

Hypotenuse with hypotenuse. A hypotenuse matched with a leg gives nothing, and such triangles cannot even be congruent (Example 1(d)).

The given leg sets the pairing. The vertex where the given leg meets the hypotenuse matches the same vertex in the other triangle, so the congruence statement is not always in alphabetical order (Example 1(b)).

HL is the named case of a wider fact. SSA data builds exactly one triangle whenever the given angle faces the longer of the two given sides (why side side angle fails, Boundaries). A right angle always does, because the hypotenuse is the longest side. An obtuse angle does too, but on this map only HL is cited as a reason in a proof.

Where this goes next

HL does its best work where right angles turn up unasked: at a distance to a line, and where a radius meets a tangent. Later, the two tangents from an outside point to a circle are shown congruent by HL, and the perpendicular from the center to a chord bisects it for the same reason as Example 4. Example 5 is one half of the angle bisector distance theorem.

As with every criterion, the triangles are rarely the goal; the parts they hand over are, which is proving a part congruent within proving triangles congruent.

Exercises

Is it HL?

\triangle ABC and \triangle DEF are right triangles with right angles at C and F. For each pair of facts, name the criterion that proves the triangles congruent (HL, SAS, ASA or AAS) and write the congruence statement, or say that none applies.

\overline{AB} \cong \overline{DE} and \overline{AC} \cong \overline{DF}
\overline{AB} \cong \overline{DE} and \overline{AC} \cong \overline{EF}
\overline{AC} \cong \overline{DF} and \overline{BC} \cong \overline{EF}
\overline{AB} \cong \overline{DE} and \angle A \cong \angle D
\overline{AC} \cong \overline{DF} and \angle A \cong \angle D
\angle A \cong \angle D and \angle B \cong \angle E

Finding lengths

\triangle PQR and \triangle STU are right triangles with right angles at R and U, PQ = ST = 25 and QR = TU = 7. Write the congruence statement and find PR and SU.
In right triangles \triangle ABC and \triangle DEF, with right angles at C and F, AB = 4x - 3, DE = 2x + 11 and BC = EF = 20. Find the value of x for which HL applies, then find AB and AC.
In right triangles \triangle ABC and \triangle DEF, with right angles at C and F, AB = 2y + 1, DE = y + 9, BC = 3z and EF = z + 10. Find y and z so that HL applies, then find AC.
Two right triangles each have a hypotenuse of 9 and a leg of 4. Find the other leg exactly, and explain why the triangles are congruent.
Two straight ramps, each 10 m long, rise from level ground to the top edge of a loading dock whose front face is vertical and 2.8 m high. How far from the face does each ramp start, and why do the two ramps meet the ground at the same angle?

Short proofs

Write each as a two-column proof, using Figure 6.

Figure 6Figures for Exercises 12 to 14.

Given. In Figure 6(a), \angle B and \angle D are right angles, C is the midpoint of \overline{BD}, and \overline{AC} \cong \overline{EC}. Prove. \triangle ABC \cong \triangle EDC.
Given. In Figure 6(b), \angle ABC and \angle ADC are right angles and \overline{AB} \cong \overline{AD}. Prove. \triangle ABC \cong \triangle ADC.
With the Given of Exercise 13, prove that \overrightarrow{CA} bisects \angle BCD.

Reasoning

HL uses a right angle, a leg and the hypotenuse. Which arrangement of three parts is that: SSS, SAS, ASA, AAS or SSA? Why does the two-triangle counterexample for that arrangement fail when the angle is a right angle?
The proof of the HL theorem ends with the SSS postulate. Show that it could end with the SAS postulate instead: which lines would SAS use, and what extra fact would it need?
Two right triangles both have a hypotenuse of 25. Must they be congruent? Give leg lengths that show your answer.
Explain why LL, two legs, needs no theorem of its own, while HL does.
Right triangles \triangle ABC and \triangle DBC share the leg \overline{BC}, have their right angles at C, lie on opposite sides of \overleftrightarrow{BC}, and have \overline{AB} \cong \overline{DB}. Explain why A, C and D lie on one line, then use the Base Angles Theorem to prove \triangle ABC \cong \triangle DBC without the Pythagorean theorem.

Answers

HL: \triangle ABC \cong \triangle DEF. The hypotenuses match, and the legs meet them at A and D.
HL: \triangle ABC \cong \triangle EDF. The leg \overline{AC} meets the hypotenuse at A and \overline{EF} meets it at E, so A \leftrightarrow E, C \leftrightarrow F, B \leftrightarrow D.
SAS: \triangle ABC \cong \triangle DEF. The right angles, congruent to each other, are included between the legs.
AAS: \triangle ABC \cong \triangle DEF. \angle A \cong \angle D, the right angles match, and \overline{AB}, \overline{DE} face the right angles.
ASA: \triangle ABC \cong \triangle DEF. \overline{AC} joins \angle A to \angle C, and \overline{DF} joins \angle D to \angle F.
None. All three pairs of angles match, which fixes the shape but not the size (why angle angle angle fails).
HL: \triangle PQR \cong \triangle STU, and PR = SU = 24. From PR^2 + 7^2 = 25^2, PR^2 = 625 - 49 = 576; corresponding parts give SU = PR.
x = 7, AB = 25 and AC = 15. HL needs AB = DE: 4x - 3 = 2x + 11, so 2x = 14. Then AB = 4(7) - 3 = 25 and AC^2 = 25^2 - 20^2 = 225.
y = 8, z = 5 and AC = 8. From 2y + 1 = y + 9, y = 8 and AB = 17. From 3z = z + 10, z = 5 and BC = 15. Then AC^2 = 17^2 - 15^2 = 64.
\displaystyle \sqrt{65}. The other leg satisfies x^2 + 4^2 = 9^2, so x^2 = 65. The triangles match in hypotenuse and a leg, so they are congruent by HL.
9.6 m. Each ramp, the dock face and the ground make a right triangle with hypotenuse 10 and leg 2.8, so the run satisfies x^2 = 10^2 - 2.8^2 = 100 - 7.84 = 92.16. The triangles are congruent by HL, so their angles at the ground are corresponding parts.
(1) \angle B and \angle D are right angles (given). (2) C is the midpoint of \overline{BD} (given). (3) \overline{BC} \cong \overline{DC} (definition of midpoint: line 2). (4) \overline{AC} \cong \overline{EC} (given). (5) \triangle ABC \cong \triangle EDC (HL theorem: lines 1, 4 and 3). \overline{AC} faces \angle B and \overline{EC} faces \angle D.
(1) \angle ABC and \angle ADC are right angles (given). (2) \overline{AB} \cong \overline{AD} (given). (3) \overline{AC} \cong \overline{AC} (reflexive property of congruence). (4) \triangle ABC \cong \triangle ADC (HL theorem: lines 1, 3 and 2). \overline{AC} faces both right angles, so it is the hypotenuse of each.
Lines (1) to (4) as in Exercise 13, then (5) \angle BCA \cong \angle DCA (corresponding parts of congruent triangles: line 4) and (6) \overrightarrow{CA} bisects \angle BCD (definition of angle bisector: line 5, \overrightarrow{CA} inside \angle BCD).
SSA: the right angle touches the leg and faces the hypotenuse. In the counterexample the swinging side lands twice on one ray, either side of the foot of the perpendicular from the hinge. With a right angle that foot is the angle’s own vertex, so the landings are on opposite rays: only one is on the ray the triangle needs, and the other gives its mirror image.
SAS would use line 3 (\overline{BC} \cong \overline{EF}), line 12 (\overline{AC} \cong \overline{DF}) and \angle C \cong \angle F, which follows from line 1 by the Right Angle Congruence Theorem. \angle C is included between \overline{AC} and \overline{BC}, and \angle F between \overline{DF} and \overline{EF}.
No. Legs 7 and 24 and legs 15 and 20 both give a hypotenuse of 25, since 7^2 + 24^2 = 625 = 15^2 + 20^2, but the legs differ. HL needs a leg to match as well.
In LL the right angle lies between the two legs, so LL is SAS. In HL it does not: the data is SSA, which is not a criterion in general, so HL needs its own argument.
\angle ACB and \angle BCD share \overrightarrow{CB}, and A, D lie on opposite sides of \overleftrightarrow{BC}, so the angles are adjacent. Their measures add to 90^\circ + 90^\circ = 180^\circ, and adjacent supplementary angles have opposite outer sides (the converse of the Linear Pair Theorem). So \overrightarrow{CA} and \overrightarrow{CD} are opposite rays: A, C, D lie on one line, C between A and D. In \triangle ABD, \overline{AB} \cong \overline{DB}, so \angle A \cong \angle D by the Base Angles Theorem. With the right angles at C and \overline{BC} \cong \overline{BC}, which faces \angle A and \angle D, AAS gives \triangle ABC \cong \triangle DBC.