Question 2In the following figure, D and E are two points on BC such that BD = DE= EC. Showthat area (ABD) = area (ADE) = area (AEC). [Remark: Note that by taking BD=DE=EC, the triangle ABC is divided into three Triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n Equal parts and joining the points of the division so obtained to the opposite vertex of BC, you can divide ΔABC into triangles of equal areas.]

Question 2In the following figure, D and E are two points on BC such t

1.	Question 2In the following figure, D and E are two points on BC such that BD = DE= EC. Showthat area (ABD) = area (ADE) = area (AEC). [Remark: Note that by taking BD=DE=EC, the triangle ABC is divided into three Triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n Equal parts and joining the points of the division so obtained to the opposite vertex of BC, you can divide ΔABC into triangles of equal areas.]
Answer» Question 2 In the following figure, D and E are two points on BC such that BD = DE= EC. Show that area (ABD) = area (ADE) = area (AEC). [Remark: Note that by taking BD=DE=EC, the triangle ABC is divided into three Triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n Equal parts and joining the points of the division so obtained to the opposite vertex of BC, you can divide $Δ A B C$ into triangles of equal areas.]