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Ex 9.3, 1 - Chapter 9 Class 9 - Areas of Parallelograms and Triangles (Deleted)

Last updated at July 18, 2019 by

Ex 9.3, 1 - In fig, E is any point on median AD of - Ex 9.3

Ex 9.3, 1 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2

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Ex 9.3, 1 In Fig, E is any point on median AD of a ABC. Show that ar (ABE) = ar (ACE). Given : ABC is a triangle with AD as medium i.e. BD = CD E is any point on AD. To Prove : ar ( ABE) = ar ( ACE) Proof: Join EB and EC Subtracting (1) and (2) ar ( ABD) ar ( EBD) = ar ( ADC) ar ( EDC) ar ( ABE) = ar ( ACE) Hence, Proved

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