Heron’s formula

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In this activity we will give two proofs of Heron’s formula.

We’ll start by giving a proof using synthetic geometry.

Part I

The bisectors of the angles of a triangle meet at a point that is the center of the triangle’s inscribed circle.

How can we prove this?

Now draw a triangle with vertices $A$ , $B$ , and $C$ . Draw the incircle. Explain why the radii of the incircle touch the sides of the triangle at right angles.

Label the intersection of the radii with $D$ between $A$ and $B$ , $E$ between $B$ and $C$ , and $F$ between $C$ and $A$ . Compute the areas of the following triangles: $\triangle AOB,\qquad \triangle BOC,\qquad \triangle COA.$ Use this to express the area of $\triangle ABC$ .

Part II

Explain why $\triangle AOD \cong \triangle AOF,\qquad \triangle BOD \cong \triangle BOE,\qquad \triangle COF \cong \triangle COE.$

If $AG \cong CE$ , explain why $|BG|$ is the semiperimeter.

Find segments in your drawing equal to the length of $s-a,\qquad s-b,\qquad s-c.$

Part III

If quadrilateral $AHBO$ has diagonals $AB$ and $OH$ with $\angle HAB$ and $\angle HOB$ being right angles, then $AHOB$ can be inscribed in a circle.

Can you prove this proposition?

The opposite angles of a cyclic quadrilateral sum to two right angles.

Can you prove this proposition?

Now we need to decorate our triangle even more:

(a): Draw $OL$ perpendicular to $OB$ cutting $AB$ at $K$ .
(b): Draw $AM$ perpendicular to $BG$ .
(c): Call the intersection of $OL$ and $AM$ , $H$ .
(d): Draw $BH$ .

Consider quadrilateral $AHBO$ , explain why opposite angles sum to two right angles.

Explain why $\triangle COF$ is similar to $\triangle BHA$ . Use this to explain why $\frac {|AB|}{|AG|} = \frac {|AH|}{r}.$

Explain why $\triangle KAH$ is similar to $\triangle KDO$ . Use this to explain why $\frac {|AK|}{|KD|} = \frac {|AH|}{r}.$

Now we see $\frac {|AB|}{|AG|} = \frac {|AK|}{|KD|}.$ Add $1$ to both sides to obtain $\frac {|BG|}{|AG|} = \frac {|AD|}{|KD|}.$

Explain why $\triangle KDO$ is similar to $\triangle ODB$ . Use this to explain why $|KD|\cdot |BD| = r^2.$

Multiply one side of $\frac {|BG|}{|AG|} = \frac {|AD|}{|KD|}$ by $\frac {|BG|}{|BG|}$ and the other side by $\frac {|BD|}{|BD|}$ to obtain $r^2|BG|^2 = |AG|\cdot |BG|\cdot |AD|\cdot |BD|.$

Explain how to deduce Heron’s formula.

A modern proof

Now give a modern proof that a high school student might give.

Which proof was harder? Why didn’t the ancient Greeks use our modern proof?