Question

the medians BE and CF of a triangle ABC intersects at G.prove that ar(triangle GBC)=area of quadrilateral AFGE.

Answer 1

Step-by-step explanation:

Given that median of a triangle divides the third side into two equal parts.

So, E and F are mid-points of sides AC and AB respectively.

Construction : Join EF.

Proof:

The line joining the mid-points of two sides of a triangle is parallel to the third side.

∴BC||EF

Triangle on the same base and between the same parallel lines are equal in area.

∴ ar (BCF) = ar (BCE)

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

⇒ar (CEG) =  ar (BFG)   .............(i)

Now, The median of a triangle divides the triangle into two triangles of equal area.

BE is median of ΔABC

∴ ar (BCE) = ar (ABE)

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

⇒ar (BCG) + ar (CEG) = ar (CEG) + ar (AFGE)   [From (i)]

⇒ ar (BCG) = ar (AFGE)

Answer 2

Heya mate
The answer of ur question is




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

Construction : Join EF.

Proof:

The line joining the mid-points of two sides of a triangle is parallel to the third side.

∴BC||EF

Triangle on the same base and between the same parallel lines are equal in area.

∴ ar (BCF) = ar (BCE)

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

⇒ar (CEG) = ar (BFG) .............(i)

Now, The median of a triangle divides the triangle into two triangles of equal area.

BE is median of ΔABC

∴ ar (BCE) = ar (ABE)

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

⇒ar (BCG) + ar (CEG) = ar (CEG) + ar (AFGE) [From (i)]

⇒ ar (BCG) = ar (AFGE)



hope it helps you