Question

1 In Fig.9.23. E is any point on median AD of a
ABC. Show that ar (ABE)=ar (ACE).
2. In a triangle ABC, E is the mid-point of median
1
AD. Show that ar (BED)=
ar(ABC)
4.
B
3
Fig. 933
3. Show that the diagonals of a parallelogram divide
it into four triangles of equal area.
4. In Fig. 9.24, ABC and ABD are two triangles on
the same base AB. If line-segment CD is bisected.
by AB at , show that ar(ABC)=ar (ABD).​

Answer 1

Answer:

nhi pata answer sorry ok

Answer 2

THIS is the first solution .

Step-by-step explanation:

proof: ar(ADB)=AR(ADC). (GIVEN)

ar(EDB)=ar(EDC)

NOW,

ar(EDB)-ar(ADB) = ar(EDC)-ar(ADC)

this implies --

ar(ABE)=ar(ACE)

HENCE PROVED.