If the diagonals of a quadrilateral divide each other proportionally prove that is a trapezium ?

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Solution

$\frac{A O}{B O} = \frac{C O}{D O}$ $[G i v e n]$
$\frac{A O}{C O} = \frac{B O}{D O}$ $. . . . . . . (1) [c r o s s m u l t i p l i c a t i o n]$
$I n$ $Δ^{l e} A B D, E O | | A B$ $(c o n s t r u c t i o n)$
$∴ \frac{A E}{E D} = \frac{B O}{D O}$ $[B y B P T] . . . . . . (2)$
$F r o m e q u a t i o n (1) a n d (2)$
$\frac{A E}{E D} = \frac{A O}{C O} (E O | | D C)$ $[C o n v e r s e o f B P T]$
$B u t$ $E O | | A B$ $(c o n s t r u c t i o n) .$
$∴ A B | | B C$
$\Rightarrow$ $I n Q u a d r i l a t e r a l$ $A B C D A B | | D C$
$∴$ $A B C D i s a t r a p e z i u m .$

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