Question

In a given figure BC is parallel to DE.Prove that: area ∆ABE =area ∆ACD

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

From the figure ,

BC //DE ,

i)

BC//DE,

Area ∆CBE = Area ∆CBD --(1)

/* Triangles on the as same base (or equal bases ) and between the same parallels are equal in area */

ii ) Now,

Area ∆ABE

= Area ∆ABC + Area ∆CBE

= Area ∆ABC + Area ∆CBD /* From (1) */

= ∆ACD

Therefore,

area ∆ABE = Area ∆ACD

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