If the medians of a `triangle`ABC intersect at G, show that
`ar(triangleAGB)=ar(triangleAGC)=ar(triangleBGC)`
`=(1)/(3)ar(triangleABC)`.

If the medians of a triangleABC intersect at G, show that ar(AGB) = ar(AGC) = ar(BGC) = 1/3 ar(ABC)

If the medians of a triangleABC intersect at G, show that ar(AGB) = ar(AGC) = ar(BGC) = 1/3 ar(ABC)

True/False If AD is the median of triangleABC and P is a point on AC such that ar(triangleADP): ar (triangleABD) = 2:3 , then ar(trianglePDC) : ar(triangleABC) =1:3

If the median of a △ABC intersect at G. show that ar (△AGC) = ar (△AGB) = ar (△BGC) = 1/3 ​ ar (△ABC)

In the figure, E is any point on median AD of a triangleABC . Show that ar( triangleABE )= ar( triangleACE ).

If ABCD is a tetrahedron such that each triangleABC, triangleABD and triangleACD has a right angle at A. If ar(triangle(ABC))=k_1, ar(triangleABD)=k_2, ar(triangleBCD)=k_3 , then ar(triangleACD) is

If ABCD is a tetrahedron such that each triangleABC, triangleABD and triangleACD has a right angle at A. If ar(triangle(ABC))=k_1, ar(triangleABD)=k_2, ar(triangleBCD)=k_3 , then ar(triangleACD) is

ABCD is a tetrahedron such that each of the triangleABC , triangleABD and triangleACD has a right angle at A. If ar(triangleABC) = k_(1). Ar(triangleABD)= k_(2), ar(triangleBCD)=k_(3) then ar(triangleACD) is