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Question -

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 AO/OC = OB/OD



Answer -

In ╬ФDOC and ╬ФBOA,

AB || CD, thus alternate interior angles willbe equal,

тИ┤тИаCDO = тИаABO

Similarly,

тИаDCO = тИаBAO

Also, for the two triangles ╬ФDOC and ╬ФBOA,vertically opposite angles will be equal;

тИ┤тИаDOC = тИаBOA

Hence, by AAA similarity criterion,

╬ФDOC ~ ╬ФBOA

Thus, the corresponding sides areproportional.

DO/BO = OC/OA

тЗТOA/OC = OB/OD

Hence, proved.

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