from a handpicked tutor in LIVE 1-to-1 classes

# If the medians of a ∆ ABC intersect at G, show that ar(AGB) = ar(AGC) = ar(BGC) = 1/3 ar(ABC)

**Solution:**

Given, ABC is a triangle

The medians of triangle ABC intersect at G

We have to show that ar(AGB) = ar(BGC) = 1/3 ar(ABC)

In triangle ABC,

Since AD is a median

ar(ABD) = ar(ACD) ------------------- (1)

In triangle BGC,

Since GD is a median

ar(GBD) = ar(GCD) ------------------ (2)

Subtracting (1) and (2),

ar(ABD) - ar(GBD) = ar(ACD) - ar(GCD)

So, ar(AGB) = ar(AGC) ------------- (3)

Similarly,

ar(AGB) = ar(BGC) ------------------ (4)

From (3) and (4),

ar(AGB) = ar(BGC) = ar(AGC) ----------------------- (5)

Now, ar(ABC) = ar(AGB) + ar(BGC) + ar(AGC)

From (5),

ar(ABC) = ar(AGB) + ar(AGB) + ar(AGB)

ar(ABC) = 3ar(AGB)

So, ar(AGB) = 1/3 ar(ABC) ----------------------------- (6)

From (5) and (6),

ar(BGC) = 1/3 ar(ABC)

ar(AGC) = 1/3 ar(ABC)

Therefore, ar(AGB) = ar(BGC) = ar(AGC) = ⅓ ar(ABC)

**✦ Try This:** If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of median AD, prove that ar(△BGC)=2 ar(△AGC)

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

**NCERT Exemplar Class 9 Maths Exercise 9.4 Problem 8**

## If the medians of a ∆ ABC intersect at G, show that ar(AGB) = ar(AGC) = ar(BGC) = 1/3 ar(ABC)

**Summary:**

Every triangle has exactly three medians, one from each vertex, and they all intersect each other at the triangle's centroid. If the medians of a ∆ ABC intersect at G, it is shown that ar(AGB) = ar(AGC) = ar(BGC) = 1/3 ar(ABC)

**☛ Related Questions:**

- 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 . . . .
- In Fig. 9.25, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn par . . . .

visual curriculum