In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that: (i) Δ AOB is similar to Δ COD. (ii) OA x OD = OB x OC.

(I) Given,

\[AO\text{ }=\text{ }2CO\text{ }and\text{ }BO\text{ }=\text{ }2DO,\]

\[AO/CO\text{ }=\text{ }2/1\text{ }=\text{ }BO/DO\]

What’s more,

\[\angle AOB\text{ }=\angle DOC\] [Vertically inverse angles]

Consequently, \[\vartriangle AOB\text{ }\sim\text{ }\vartriangle COD\][SAS model for similarity]

(ii) As, \[AO/CO\text{ }=\text{ }2/1\text{ }=\text{ }BO/DO\][Given]

Consequently,

\[OA\text{ }x\text{ }OD\text{ }=\text{ }OB\text{ }x\text{ }OC\]