Question

In ΔAВC; E and F are the mid-points of sides AB and AC respectively. If FB and CE intersect at point ‘O’, prove that area (ΔOBC) = area (◻AEOF).

Answer 1

Given:

In ΔABC, E and F are the mid-points of AB and AC, respectively.

FB and CE meet at O.

To Prove: area (ΔOBC) = area (◻AEOF).

Proof [Step-wise]:

1. Since E and F are mid-points of AB and AC, EF || BC mid‑segment theorem. 

Also, FB and CE are medians of ΔABC.

So, their intersection O is the centroid of ΔABC intersection of medians.

2. As O is the centroid, it divides each median in the ratio 2 : 1 measured from the vertex.

In particular, on median CE, we have CO : OE = 2 : 1 and on median BF, we have BO : OF = 2 : 1.

3. Consider triangle BCE. Point O lies on CE, so the areas of triangles BEO and BCO are in the ratio of the corresponding bases EO and OC; they have the same altitude from B to line CE. 

Hence, area (BEO) : area (BCO)

= EO : OC

= 1 : 2

Thus, area (BEO) = `1/2` × area (BCO). 

Therefore, area (BCO) = 2 × area (BEO).   ...(1)

4. Similarly, in triangle BCF, point O lies on BF.

So, area (CFO) : area (CBO)

= FO : OB

= 1 : 2

Giving area (CFO) = `1/2` × area (CBO)

= `1/2` × area (BCO)

Hence, area (CFO) = area (BEO).   ...(2)

5. Now compare triangles with bases on AB and AC:

Triangles AEO and BEO have the same altitude from O to line AB, and AE = EB (E is midpoint).

So, area (AEO) = area (BEO).

Triangles AFO and CFO have the same altitude from O to line AC, and AF = FC (F is midpoint).

So, area (AFO) = area (CFO).

6. Compute area of quadrilateral AEOF:

area (◻AEOF) = area (AEO) + area(AFO)

= area (BEO) + area(CFO)   ...(By step 5) 

= area (BEO) + area (BEO)   ...(By (2)) 

= 2 × area (BEO) = area (BCO)   ...(By (1) = area (ΔOBC).

Therefore, area (ΔOBC) = area (◻AEOF), as required.