3 Tutor System
Starting just at 265/hour

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

In $$\triangle$$DOC and $$\triangle$$BOA,

AB || CD, thus alternate interior angles will be equal,
$$\angle$$CDO = $$\angle$$ABO

Similarly,

$$\angle$$DCO = $$\angle$$BAO

Also, for the two triangles $$\triangle$$DOC and $$\triangle$$BOA, vertically opposite angles will be equal;
$$\angle$$DOC = $$\angle$$BOA
Hence, by AAA similarity criterion,
$$\triangle$$DOC ~ $$\triangle$$BOA

Thus, the corresponding sides are proportional.
DO/BO = OC/OA
OA/OC = OB/OD

Hence, proved.