Question

In the adjoining figure, D is the mid-point of BC and E be any point on AD.

Prove that : 

1. area (ΔEBD) = area (ΔECD).
2. area (ΔABE) = area (ΔACE).

Answer 1

Given: D is the midpoint of BC. E is any point on AD, so E lies on segment AD.

To Prove:

1. area (ΔEBD) = area (ΔECD). 
2. area (ΔABE) = area (ΔACE).

Proof [Step-wise]:

Part (i)

1. BD = DC because D is the midpoint of BC.   ...(Given)

2. Triangles EBD and ECD have their bases BD and DC on the same straight line BC and share the same vertex E; hence, they have the same altitude from E to line BC.   ...(Triangles with a common vertex and bases on the same line have areas proportional to their bases).

3. Area (ΔEBD) : Area (ΔECD) = BD : DC   ...(Areas are proportional to corresponding bases when the vertex is the same).

4. Since BD = DC, the ratio is 1. So, Area (ΔEBD) = Area (ΔECD).

Part (ii)

1. Write area (ΔABE) as area (ΔABD) – area (ΔEBD).   ...(Because E lies on AD, so ΔABE is the part of ΔABD remaining after removing ΔEBD).

2. Similarly, area (ΔACE) = area (ΔACD) – area (ΔECD).

3. Since D is the midpoint of BC, triangles ABD and ACD are on the same base AD.

More directly: ABD and ACD have equal bases BD and DC on BC and the same vertex A.

So, area (ΔABD) = area( ΔACD).

4. From part (i), we have area (ΔEBD) = area (ΔECD).

5. Therefore, area (ΔABE)

= area (ΔABD) – area (ΔEBD)

= area (ΔACD) – area (ΔECD) 

= area (ΔACE)