The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆ GBC = area of the quadrilateral AFGE.

Solution:

Given, ABC is a triangle

The medians BE and CF intersect at G

We have to prove that ar(GBC) = ar(AFGE)

We know that the median of a triangle divides it into two triangles of equal areas.

Since BE is the median

So, ar(ABE) = ar(CBE)

ar(ABE) = 1/2 ar(ABC) --------------- (1)

Since CF is the median

So, ar(ACF) = ar(BCF)

ar(BCF) = 1/2 ar(ABC) --------------- (2)

From (1) and (2),

ar(ABE) = ar(BCF) ------------------ (3)

On subtracting ar(GBF) from both sides in (3),

ar(ABE) - ar(GBF) = ar(BCF) - ar(GBF)

From the figure,

ar(ABE) - ar(GBF) = ar(AFGE)

ar(BCF) - ar(GBF) = ar(GBC)

Therefore, ar(AFGE) = ar(GBC)

✦ Try This: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 9

---

NCERT Exemplar Class 9 Maths Exercise 9.4 Problem 3

The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆ GBC = area of the quadrilateral AFGE.

Summary:

The medians BE and CF of a triangle ABC intersect at G. It is proven that the area of ∆ GBC = area of the quadrilateral AFGE

---

☛ Related Questions:

• In Fig. 9.24, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY)
• ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the m . . . .
• In ∆ ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LO . . . .