
Get live Maths 1-on-1 Classs - Class 6 to 12
Ex 9.3
Ex 9.3, 2 Important Deleted for CBSE Board 2023 Exams
Ex 9.3, 3 Deleted for CBSE Board 2023 Exams
Ex 9.3, 4 Deleted for CBSE Board 2023 Exams
Ex 9.3, 5 Important Deleted for CBSE Board 2023 Exams
Ex 9.3, 6 Important Deleted for CBSE Board 2023 Exams
Ex 9.3, 7 Important Deleted for CBSE Board 2023 Exams
Ex 9.3, 8 Deleted for CBSE Board 2023 Exams
Ex 9.3, 9 Important Deleted for CBSE Board 2023 Exams
Ex 9.3, 10 Deleted for CBSE Board 2023 Exams
Ex 9.3, 11 Important Deleted for CBSE Board 2023 Exams
Ex 9.3, 12 Important Deleted for CBSE Board 2023 Exams
Ex 9.3, 13 Deleted for CBSE Board 2023 Exams
Ex 9.3, 14 Deleted for CBSE Board 2023 Exams
Ex 9.3, 15 Deleted for CBSE Board 2023 Exams
Ex 9.3, 16 Important Deleted for CBSE Board 2023 Exams
Last updated at March 16, 2023 by Teachoo
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