Question 540938
Equilateral triangles are very symmetrical. In an equilateral triangle, all three sides have the same length and all 3 angles have the same measure (60 degrees).
{{{drawing( 250, 150, -10, 10, -2, 10,
  locate( 0, 0, P ),
  locate( -5, 0, A ), locate( 5, 0, B ), locate( 0, 10, C ),
  blue(line( 5, 0, -5, 0 )),
  line( 5, 0, 0, 8.66 ),
  line( -5, 0, 0, 8.66),
  red(line( 0, 0, 0, 8.66 ))
  )}}}Being a symmetrical figure, the CP altitude bisects the base AB.
So the length of each triangle side (AB, BC, and CA) is 8.
The length of segment PB is half of the base length, so it's 4.
Right triangle PBC has a hypotenuse (BC) of length 8, and one leg of length 4.
We use Pythagoras theorem to find the length of the other leg (altitude CP).
Let's call that length x.
{{{x^2+4^2=8^2}}} --> {{{x^2+16=64}}} --> {{{x^2=64-16}}} --> {{{x^2=48}}}
Since the length of a segment must be a positive number, never a negative one, there is only one solution:
{{{x=sqrt(48)=sqrt(16*3)=sqrt(16)*sqrt(3)=4sqrt(3)}}}