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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 ΔDOC and ΔBOA,

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

∴∠CDO = ABO

Similarly,

DCO = BAO

Also, for the two triangles ΔDOC and ΔBOA,vertically opposite angles will be equal;

∴∠DOC = BOA

Hence, by AAA similarity criterion,

ΔDOC ~ ΔBOA

Thus, the corresponding sides areproportional.

DO/BO = OC/OA

OA/OC = OB/OD

Hence, proved.

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