Check sibling questions

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

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

Solve all your doubts with Teachoo Black (new monthly pack available now!)

Join Teachoo Black

Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.