# The medians BE and CF of ΔABC intersects at G. Prove that ar (ΔGBC) = ar (quad AFGE).

2
by pravin3103

2014-12-28T12:35:59+05:30

Given median of a triangle divides the third side into two equal parts. So E and F are the mid points of side AC and AB respectively.

Const: JOIN EF

Proof ; As we know that the line joining the mid points of 2 sides of a triangle is // to the third side

therefore BC//EF

Triangles on the same base and between same // lines are equal in area.

therefore, ar(BCF)= ar(BCE)

implies ar(BCG)+ ar(CEG) = ar(BCG) + ar(BFG)

implies ar(CEG)= ar(BFG) 1)

Now the median of triangle divides the triangle into 2 equal areas

BE is the median of triangle ABC

therefore ar(BCE)=ar(ABE)

implies ar(BCG) + ar(CEG)= ar(BFG) + ar(AFGE)

implies ar(BCG) + ar(CEG)= ar(CEG) + ar(AFGE) from 1

hence ar(BCG) = ar(AFGE)

2014-12-28T13:34:35+05:30

Given ,

Median of a triangle divides the third side into two equal parts.

E and F are the mid points of side AC and AB.

we know that the line joining the mid points of 2 sides of a triangle is paralell to the third sid.

so,BC is parallel to EF.

Triangles with the same base and which are included  between the two parallel lines are equal in area.

so, ar(BCF)= ar(BCE)

ar(BCG)+ ar(CEG) = ar(BCG) + ar(BFG)

ar(CEG)= ar(BFG) 1)

median of triangle divides the triangle into 2 equal area.

BE is the median of triangle ABC.

so,ar(BCE)=ar(ABE)

ar(BCG) + ar(CEG)= ar(BFG) + ar(AFGE)

ar(BCG) + ar(CEG)= ar(CEG) + ar(AFGE) from 1

HENCE ar(BCG) = ar(AFGE)