Proof:

Let О be the point of intersection of segments АВ and СD, where АО = ОD and СО = ОВ.

Since АО = ОD, we can say that АО and ОD are equal in length and form one side of the triangle АОС.

Similarly, СО = ОВ means that СО and ОВ are equal in length and form one side of the triangle DОВ.

Now, by the Side-Angle-Side (SAS) congruence theorem, we can conclude that АОС and DОВ are congruent triangles.

Therefore, all corresponding angles of АОС and DОВ are equal, including angle АОС = angle DОВ.

This means that АОС and DОВ are also similar triangles. And since corresponding angles of similar triangles are equal, we can say that АОС = DОВ.

Hence, we have proved that АОС = DОВ.

Q.E.D.

Дано:

• AB и CD - две отрезка, которые пересекаются в точке O
• AO = OD
• CO = OB

Требуется доказать:

AОS = DОВ