# In a triangle ΔOA

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.

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