**Download NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.3 – Areas of Parallelograms And Triangles**

**1. In given figure, E is any point on median AD of a ∆ ABC. Show that
**

**ar (ABE) = ar (ACE).**

**Solution :**

Given that, AD is the median of ∆ ABC, BD = DC

=> ar (∆ ABD ) = ar (∆ ADC )

=> ar (∆ ABE ) + ar (∆ EBD ) = ar (∆ ACE ) + ar (∆ ECD ) ….(1)

It can be clearly observed that, in the ∆ EBC, ED is the median.

=> ar ( ∆ EBD ) = ar ( ∆ ECD ) ….(2)

On subtracting equation 2 from equation 1, we get

ar ( ∆ ABE ) = ar ( ∆ ACE )

**Hence Proved.**

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