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.
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
\[ \triangle AOD \cong \triangle AOF,\qquad \triangle BOD \cong \triangle BOE,\qquad \triangle COF \cong \triangle COE. \]
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.
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 \[ r^2|BG|^2 = |AG|\cdot |BG|\cdot |AD|\cdot |BD|. \]