ABCD is a parallelogram and P is the midpoint of the side `bar(BC)`. Prove that `bar(AC) " and " bar(DP)` meet in a common point of trisection.

ABCD is a parallelogram, P and Q are the mid-points of the sides bar(AB) " and " bar(DC) respectively. Show that bar(DP) " and " bar(BQ) trisect bar(AC) and are trisected by bar(AC) .

ABCD is a parallelogram and P, Q are the mid points of bar(AD) and bar(BC) respectively. Show that : bar(BD) and bar(PQ) bisect each other.

ABCD is parallelogram.If L and M are the middle points of BC and CD, then bar(AL)+bar(AM) equals

ABCD is parallelogram. If L and M are the middle points of BC and CD, then bar(AL)+bar(AM) equals

If D is the mid -point of side AB of DeltaABC , then bar(AB) + bar(BC) + bar(AC) =

If D is the mid -point of side AB of DeltaABC , then bar(AB) + bar(BC) + bar(AC) =

ABCD is a parallelogram and P is the mid-point of bar(DC) . If Q is a point on bar(AP) , such that bar(AQ)=2/3bar(AP) , show that Q lies on the diagonal bar(BD) " and " bar(BQ)=2/3bar(BD) .

If ABCD is a parallelogram then bar(AC)+bar(BD)=

If D is the midpoint of the side BC of a triangle ABC then bar(AB)^(2)+bar(AC)^(2)=