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  • Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA by OC equals OB by OD

Q3  Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity  criterion for two triangles, show that \frac{OA }{OC} = \frac{OB }{OD }

 

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In  \triangle DOC\, and\, \triangle BOA , we have

\angle CDO=\angle ABO        ( Alternate interior angles as AB||CD)

\angle DCO=\angle BAO        ( Alternate interior angles as AB||CD)

\angle DOC=\angle BOA        ( Vertically opposite angles are equal)

\therefore \triangle DOC\, \sim \, \triangle BOA    ( By AAA)

\therefore \frac{DO}{BO}=\frac{OC}{OA}      ( corresponding sides are equal)

\Rightarrow \frac{OA }{OC} = \frac{OB }{OD }

Hence proved 

 

 

 

 

 

 

Posted by

seema garhwal

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