In a triangle ΔOAB,
. If P and Q are points of trisection of AB, prove that 
Given:-
, P and Q are trisection of AB
i.e. AP = PQ = QB or 1:1:1 division of line AB
To Prove:-![]()

Proof:- Let
be position vector of O, A and B respectively
Now, Find position vector of P, we use section formulae of internal division: Theorem given below
“Let A and B be two points with position vectors ![]()
respectively, and c be a point dividing AB internally in the ration m:n. Then the position vector of c is given by ![]()
By above theorem, here P point divides AB in 1:2, so we get
⇒ ![]()
⇒ ![]()
Similarly, Position vector of Q is calculated
By above theorem, here Q point divides AB in 2:1, so we get
⇒ ![]()
⇒ ![]()
Length OA and OB in vector form
⇒ ![]()
⇒ ![]()
⇒ ![]()
⇒ ![]()
Now length/distance OP in vector form
![]()
⇒ ![]()
⇒ ![]()
⇒ ![]()
⇒ ![]()
![]()
⇒ ![]()
length/distance OQ in vector form
![]()
⇒ ![]()
⇒ ![]()
⇒ ![]()
⇒ ![]()
![]()
⇒ ![]()
Taking LHS
OP2 + OQ2
= ![]()
= ![]()
as we know in case of dot product
![]()
![]()
Angle between OA and OB is 90°,
⇒ ![]()
⇒ ![]()
Therefore, OP2 + OQ2
= ![]()
= ![]()
= ![]()
= ![]()
As from figure OA2 + OB2 = AB2
= ![]()
= RHS
Hence, Proved.