Question

Pls ans this as soon as possible i will Mark you brainliest

Answer 1

it is very easy . i am knowing how to solve this
In ΔBECΔBEC , DD is a midpoint so EDED is a median.[Math Processing Error]
∴\arBDE=\arCDE(1)(1)∴\arBDE=\arCDE
Now, ΔADEΔADE and ΔBDEΔBDE are on the same base DEDEand between the same parallels DEDE and ABAB.
∴\arADE=\arBDE(2)(2)∴\arADE=\arBDE
From (1)(1) and (2)(2):
\arADE=\arCDE(3)(3)\arADE=\arCDE
i.e. DEDE divides ΔADCΔADC into two parts of equal area.
∴∴ DEDE is a median and EE is the midpoint of AC.AC.
■