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 $$OA\over OC$$ = $$OB\over OD$$.

Solution :

Given : A trapezium ABC whose diagonals AC and BD intersect each other at O and AB || DC.

To Prove : $$OA\over OC$$ = $$OB\over OD$$

Proof : In triangle DOC and AOB,

AB || DC and AC is transversal, then

$$\angle$$ DCO = $$\angle$$ OAB                     (Alternate angles)

$$\angle$$ ODC = $$\angle$$ OBA                    (Alternate angles)

$$\angle$$ DOC = $$\angle$$ AOB                       (vertically opposite angles)

Hence, By AAA similarity,

$$\triangle$$ DOC ~ $$\triangle$$ BOA

$$\implies$$    $$OA\over OC$$ = $$OB\over OD$$