$$\begin{gathered} 14. In the adjoining figure , ABCD is a trapazium with AD and BC as parallel sides and\angle BAD ={90}^o \\ BD and AC cut at E. Prove that the triangle ABE is equal in area to the triangle CED \end{gathered}$$

Dear Student,

Please find below the solution to the asked query:

13 ) From given diagram we know : BC | | DE 

So,

Height of $∆$ BCD =  Height of $∆$ BCE  =  h     (  As both triangles lies between same parallel lines " DE | |BC " by taking base ' BC ' )

We know :  Area of triangle  = $\frac{1}{2}$ $\times$ Base $\times$ Height , So

Area of $∆$ BCD = $\frac{1}{2}$ $\times$ BC $\times$ h                                            --- ( 1 )

And

Area of $∆$ BCE  = $\frac{1}{2}$ $\times$ BC $\times$ h                                            --- ( 2 )

From equation 1 and 2 we get :

Area of $∆$ BCD = Area of $∆$ BCE

Now we subtract ' Area of $∆$ BOC ' on both hand side we get :

Area of $∆$ BCD - Area of $∆$ BOC =  Area of $∆$ BCE - Area of $∆$ BOC

We know from Third Axiom of Euclid " If equals be subtracted from equals, the remainders are equal. " So

Area of $∆$ BOD =  Area of $∆$ COE                                                          ( Hence proved )


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