Heron's Formula

Introduction

Let $a, b, c$ denote the side lengths of a triangle $ABC$ opposite to vertices $A, B, C$ respectively. Note that three positive numbers $a, b, c$ are the side lengths of a triangle only if they satisfy the triangle inequalities $a \lt b + c$, $b \lt c + a$, $c \lt a + b$; otherwise one of $s-a,\ s-b,\ s-c$ below is $0$ or negative, and the quantity under the square root in the formula is not positive. Define the semiperimeter $s$ as

Then the area $S$ of triangle $ABC$ is given by

This formula computes the area from the three side lengths alone, making it extremely useful when coordinates or angles are unknown. It was presented by Heron of Alexandria in his work Metrica in the 1st century AD.

Proof 1: Algebraic Proof Using the Law of Cosines

Drop a perpendicular of length $h$ from $A$ to the line $BC$ (Figure 2). Figure 2 shows the case where the foot lies on the side $BC$; if $B$ or $C$ is obtuse, the foot lies on the extension of $BC$. In either case $h$ is the height relative to the base $BC$, and the equations below hold unchanged. Since the area of a triangle equals base times height divided by 2 (proof),

we can write

The altitude $h$ can be expressed using the angle $C$:

where we used $\sin C \gt 0$ since $0 \lt C \lt \pi$. The law of cosines gives

so that

Let $P = (a+b)^2 - c^2$ and $Q = c^2 - (a-b)^2$. Bringing to a common denominator,

(since $2ab + a^2 + b^2 - c^2 = (a+b)^2 - c^2$ and $2ab - a^2 - b^2 + c^2 = c^2 - (a-b)^2$). Substituting these into (6),

Substituting (9) into (3),

Both $P$ and $Q$ are differences of squares, so they factor; using $a+b+c = 2s$ from (1),

Hence $PQ = 16\,s(s-a)(s-b)(s-c)$, and by (10),

This is Heron's formula (2). $\blacksquare$

Proof 2: Visual Proof Using the Incircle

The following elegant proof, due to R. B. Nelsen (2001), illustrates each step with diagrams. It derives Heron's formula via two lemmas.

Lemma 1: $S = rs$

Let $I$ be the center and $r$ the radius of the incircle of the triangle (Figure 3). The angle bisectors from each vertex pass through $I$. Dropping perpendiculars from $I$ to each side partitions the triangle into six right triangles (Figure 4).

Since two tangent segments from an external point to a circle have equal length, the tangent lengths from each vertex are:

• From vertex $A$: $x$ (on both sides $AB$ and $CA$)
• From vertex $B$: $y$ (on both sides $AB$ and $BC$)
• From vertex $C$: $z$ (on both sides $BC$ and $CA$)

From the side lengths we obtain the system of equations:

Adding (i)+(ii)+(iii) gives $2(x+y+z) = a+b+c = 2s$, so

Subtracting (i), (ii), (iii) from (iv) respectively,

That is,

The six right triangles form three congruent pairs with legs $(r, x), (r, x), (r, y), (r, y), (r, z), (r, z)$.

Rearranging these six right triangles, they tile a rectangle of height $r$ and width $x + y + z = s$ (Figure 5).

Lemma 2: $xyz = r^2 s$

In the six right triangles of Figure 4, let $\alpha, \beta, \gamma$ be the half-angles at each vertex created by the angle bisectors. Since the interior angles of a triangle sum to $\pi$,

From each right triangle (Figure 6),

We now prove the following identity for positive angles satisfying (14):

Proof of (16): From (14), $\gamma = \dfrac{\pi}{2} - (\alpha + \beta)$. The complementary angle relation $\tan(\pi/2 - \theta) = 1/\tan\theta$ gives

Applying the addition formula $\tan(\alpha+\beta) = \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ and taking the reciprocal,

Multiplying both sides of (17b) by $\tan\alpha+\tan\beta$ gives

Substituting this into the left-hand side of (16), $\tan\alpha\tan\beta + (\tan\alpha+\tan\beta)\tan\gamma$, gives $\tan\alpha\tan\beta + (1-\tan\alpha\tan\beta) = 1$. $\blacksquare$

Substituting (15) into (16),

Since $x + y + z = (s-a)+(s-b)+(s-c) = s$,

Main Proof

Squaring Lemma 1 ($S = rs$) gives $S^2 = r^2 s^2$. Substituting Lemma 2 ($xyz = r^2 s$),

Therefore $S = \sqrt{s(s-a)(s-b)(s-c)}$, completing the proof of Heron's formula. $\blacksquare$

Proof 3: The Geometric Proof Handed Down in Metrica I.8

Finally, we present the proof handed down in Heron's Metrica I.8, following Heath's translation and commentary. The formula also appears in Heron's Dioptra, and some Arabic sources attribute it to Archimedes.

Up to Lemma 1 ($S = rs$) and the tangent lengths $s-a,\ s-b,\ s-c$ (Eq. (12)), the argument is the same as in Proof 2. The difference comes after that: Proof 2 obtains $xyz = r^2 s$ from the tangent addition formula, whereas this proof uses no trigonometric functions. It draws auxiliary lines and reaches the result using only a cyclic quadrilateral and ratios of similar triangles.

Construction

Let $O$ be the center of the incircle and $D,\ E,\ F$ the points where it touches $BC,\ CA,\ AB$ (Figure 7).

1. Extend $CB$ beyond $B$ to a point $H$ with $BH = AF$.
2. Draw the line through $O$ perpendicular to $OC$, meeting $BC$ at $K$.
3. Erect the perpendicular to $BC$ at $B$, meeting the line of step 2 at $L$. Join $C$ and $L$.

Step 1: The area in terms of $CH$ and $OD$

The two tangents from each vertex are equal, so $BD = BF$, $CD = CE$, $AF = AE$. Hence

(adding each vertex's tangent length once gives the semiperimeter). By Lemma 1,

Step 2: A cyclic quadrilateral

Since $OL \perp OC$, $\angle COL = 90^\circ$, and since $BL \perp BC$, $\angle CBL = 90^\circ$. Hence both $O$ and $B$ lie on the circle with diameter $CL$, and the quadrilateral $COBL$ is inscribed in it. Opposite angles of a cyclic quadrilateral sum to $180^\circ$, so

On the other hand, the six angles around $O$ are equal in pairs, $\angle AOF = \angle AOE$, $\angle BOF = \angle BOD$, $\angle COD = \angle COE$, and they sum to $360^\circ$. Hence $\angle AOF + \angle BOD + \angle COD = 180^\circ$, that is,

Together these give $\angle AOF = \angle CLB$. The right triangles $AOF$ (right angle at $F$) and $CLB$ (right angle at $B$) are therefore similar, and

(using $AF = BH$ and $FO = OD = r$).

Step 3: A second pair of similar triangles

The right triangles $KDO$ (right angle at $D$) and $KBL$ (right angle at $B$) have equal vertical angles at $K$, so they are similar (Figure 8). Hence

Rewriting Eq. (21) as $\dfrac{CB}{BH} = \dfrac{BL}{OD}$ and combining it with Eq. (22) gives

Adding $1$ to both sides, and using $CB + BH = CH$ and $BK + KD = BD$,

Step 4: The altitude to the hypotenuse

The triangle $COK$ has a right angle at $O$, and $OD$ is the altitude to the hypotenuse $CK$. The triangles $CDO$ and $ODK$ are similar, so $CD : OD = OD : DK$, that is,

Step 5: Putting it together

Multiplying the numerator and denominator of the left side of Eq. (23) by $CH$, those of the right side by $DC$, and using Eq. (24),

Clearing denominators gives $CH^2 \cdot OD^2 = CH \cdot HB \cdot BD \cdot DC$, and the left side is $S^2$ by Eq. (20). Since $CH = s$, $HB = AF = s-a$, $BD = s-b$, $DC = s-c$,

$\blacksquare$

In Step 3 we used $BK + KD = BD$ because $K$ always lies between $D$ and $B$, which can also be checked with angles alone. Since $\angle COK = 90^\circ \gt \angle COD$, $K$ lies on the opposite side of $D$ from $C$. In the right triangle $COK$, $OD$ is the altitude to the hypotenuse, so $\angle DOK = \angle OCD = \dfrac{C}{2}$; in the right triangle $ODB$, $\angle DOB = 90^\circ - \dfrac{B}{2}$. Since $\dfrac{B}{2} + \dfrac{C}{2} \lt 90^\circ$, we get $\angle DOK \lt \angle DOB$, so $K$ lies between $D$ and $B$. The triangle in Figure 7 (sides $7, 8, 9$) is the example used in Metrica I.8; its area is $\sqrt{12 \cdot 5 \cdot 4 \cdot 3} = \sqrt{720}$.

Worked Example

Consider the right triangle with $a = 4,\ b = 5,\ c = 3$ (Figure 9).

The semiperimeter is

Applying Heron's formula,

This agrees with base $4 \times$ height $3 \div 2 = 6$.

References

• Heron's formula — Wikipedia
• T. L. Heath, A History of Greek Mathematics, Vol. II, Oxford (1921), pp. 321–323 (translation of and commentary on the proof in Metrica I.8). [TIB full text]
• R. B. Nelsen, "Heron's Formula via Proofs Without Words," The College Mathematics Journal, Vol. 32, No. 4 (2001), pp. 290–292. [JSTOR]
• Heron's formula — Wikipedia (Japanese)

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