# In ∆ ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC).

**Solution:**

Given, ABC is a triangle

L and M are the points on AB and AC

LM || BC

We have to prove that ar(LOB) = ar(MOC)

We know that the area of triangles on the same base and between the same parallel lines are equal.

Triangles LBC and MBC are on the same base BC and between the same parallel lines BC and LM.

So, ar(LBC) = ar(MBC)

From the figure,

ar(LBC) = ar(LOB) + ar(BOC)

Also, ar(MBC) = ar(MOC) + ar(BOC)

Now, ar(LOB) + ar(BOC) = ar(MOC) + ar(BOC)

Cancelling out common terms,

ar(LOB) = ar(MOC)

Therefore, it is proved that ar(LOB) = ar(MOC)

**✦ 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 6**

## In ∆ ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC).

**Summary:**

The midpoint is the middle point of a line segment. It is equidistant from both endpoints. In ∆ ABC, if L and M are the points on AB and AC, respectively such that LM || BC. It is proven that ar (LOB) = ar (MOC)

**☛ Related Questions:**

- In Fig. 9.25, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn par . . . .
- If the medians of a ∆ ABC intersect at G, show that ar(AGB) = ar(AGC) = ar(BGC) = 1/3 ar(ABC)
- In Fig. 9.26, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are str . . . .

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