Viviani's Theorem
Theorem
Let $T$ be an equilateral triangle.
Let $P$ be a point inside $T$.
Let $x$, $y$ and $z$ be the lengths of the perpendiculars dropped from $P$ to each of the three sides of $T$.
Then;
- $x + y + z = h$
where $h$ is the height of $T$.
Proof 1
Let $T = \triangle ABC$ be an equilateral triangle whose vertices are $A$, $B$ and $C$.
Let $h$ be the height of $T$.
Let $a$ be the length of one side of $T$.
Let $P$ be a point inside $T$.
Let $\triangle APB$, $\triangle BPC$ and $\triangle CPA$ be the three triangles formed by joining $P$ to each of the three vertices $A$, $B$ and $C$ of $T$.
Let the heights of $\triangle APB$, $\triangle BPC$ and $\triangle CPA$ be $x$, $y$ and $z$.
By definition, these heights are the lengths of the perpendiculars dropped from $P$ to each of the three sides of $T$.
Let $A$ be the area of $T$.
By Area of Triangle in Terms of Side and Altitude:
- $A = \dfrac {a h} 2$
But we also have that the area of $T$ is also equal to the sum of the areas of each of $\triangle APB$, $\triangle BPC$ and $\triangle CPA$.
By Area of Triangle in Terms of Side and Altitude, these areas are equal to $\dfrac {a x} 2$, $\dfrac {a y} 2$ and $\dfrac {a z} 2$.
That is:
- $A = \dfrac {a h} 2 = \dfrac {a x} 2 + \dfrac {a y} 2 + \dfrac {a z} 2$
from which it follows that:
- $h = x + y + z$
$\blacksquare$
Proof 2
Let $T = \triangle ABC$ be an equilateral triangle whose vertices are $A$, $B$ and $C$.
Let $h$ be the height of $T$.
Let $P$ be a point inside $T$.
Let $\triangle PDE$, $\triangle PFG$ and $\triangle PJH$ be three equilateral triangles constructed from $P$ to each side of $ABC$ as depicted.
Let the heights of $\triangle PDE$, $\triangle PFG$ and $\triangle PJH$ be $x$, $y$ and $z$.
Let $\triangle CGK$ be an equilateral triangle constructed also as depicted.
The result follows by inspection.
$\blacksquare$
Source of Name
This entry was named for Vincenzo Viviani.