Question

If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA² + PB² + PC² = GA² + GB² + GC² + 3 GP².

Answer 1

Hence it is proved that, PA² + PB² + PC² = GA² + GB² + GC² + 3 GP²

Given,

In a triangle ABC,

let the coordinates of vertices of triangle be: A(x1, y1), B(x2, y2) and C(x3, y3)

and the coordinates of the centroid be: G (u, v)

u = (x1 + x2 + x3) / 3, and v = (y1 + y2 + y3) / 3

and the coordinates of P(h, k)

To prove,

PA² + PB² + PC² = GA² + GB² + GC² + 3 GP²

Proof:

LHS:

PA² + PB² + PC²

= (h-x1)² + (k-y1)² + (h-x2)² + (k-y2)² + (h-x3)²+(k-y3)²

= 3(h²+k²) + (x1² + x2² + x3²)+(y1² + y2² + y3²)-2h(x1+x2+x3)-2k(y1+y2+y3)

= 3(h² + k²)+(x1² + x2² + x3²)+(y1² + y2² + y3²)-2h(3u)-2k(3v)

RHS:

GA²+GB²+GC²+3GP²

= (u-x1)²+(v-y1)²(u-x2)²+(v-y2)²+(u-x3)²+(v-y3)²+3[(u-h)²+(v-k)²]

= 3(u² + v²)+(x1² + y1² + x2² + y2² + x3² + y3²) - 2u(x1+x2+x3) - 2v(y1+y2+y3) + 3[u² + h² - 2uh + v² + k² - 2vk]

= 6(u² + v²)+(x1² + y1² + x2² + y2² + x3² + y3²) - 2u(3u) - 2v(3v) + 3(h² + k²) -6uh-6vk

= 3(h² + k²)+(x1² + x2² + x3²)+(y1² + y2² + y3²)-2h(3u)-2k(3v)

as, LHS = RHS

Hence the proof.