question_answer

                                          Diagonals AC and BD of a trapezium ABCD with \[AB\parallel DC\]intersect each other the point O. Then:

A)  \[\frac{OA}{OC}=\frac{OB}{OD}\]      

B)  \[\frac{AB}{OC}=\frac{OA}{OC}\]

C)  \[\angle OAB=\angle ODC\]       

D)  \[\frac{OA}{OB}=\frac{OC}{OD}\]

Correct Answer: A

Solution :

(a): Draw ABCD is a trapezium and AC and BD are diagonals intersect at O. In figure, \[AB\parallel DC\]          (Given) \[\Rightarrow \]\[\angle 1=\angle 3,\angle 2=\angle 4\]         (Alternate interior angles) Also, \[\angle DOC=\angle BOA\]             (Vertically opposite angles) \[\Rightarrow \]\[\Delta \,OCD\tilde{\ }\Delta \,OAB\]           (Similar triangle) \[\Rightarrow \]\[\frac{OC}{OA}=\frac{OD}{OB}\]                       (Ratio of the corresponding sides of the similar triangles) \[\Rightarrow \]\[\frac{OA}{OC}=\frac{OB}{OD}\]                       (Taking reciprocals) Hence proved.