ABC is any triangle. P is the mid-point of BC,Q is the mid-point of CAand R is the mid-point of AB. The expression `vec(AP) +vec(BQ) +vec(CR)` equals

In the figure, M is the mid-point of [AB] and N is the mid-point of [CD] and O is the mid-point of [MN]. Prove that : vec(BC)+vec(AD)=2vec(MN) .

In the figure, M is the mid - point of [AB] and N is the mid - point of [CD] and O is the mid - point of [MN]. Prove that : (i) vec(OA)+vec(OB)+vec(OC)+vec(OD)=vec(0) (ii) vec(BC)+vec(AD)=2vec(MN) .

In figure,P is the mid-point of BC, Q is the mid-point of AP, such that BQ produced meets AC at R .Prove that 3RA=CA

If ABC is triangle and D is the mid-point of [BC], prove that vec AB + vec AC= 2 vec AD .

D is the mid-point of side AB of the triangle ABC.E is the mid-point of CD and F is the mid-point of AE. Prove that 8 xx " area of " (DeltaAFD) = " area of " Delta ABC

D is the mid-point of side AB of the triangle ABC.E is the mid-point of CD and F is the mid-point of AE. Prove that 8 xx " area of " (DeltaAFD) = " area of " Delta ABC

In Delta ABC , P is a point on BC such that BP:PC=4:5 and Q is the mid - point of BP. Then area (Delta ABQ) : area (Delta ABC) is equal to :

In triangle ABC, M is mid-point of AB, N is mid-point of AC and D is any point in base BC. Use Intercept Theorem to show that MN bisects AD.