Equal segments AB and CD meet at point O, with AC || BD. Prove that triangle BOD is isosceles.

Since, by condition, AC is parallel to BD, the triangles AOC and BOD are similar in two angles. Angle AOC = BOD as vertical angles, angle OAC = OBD as criss-crossing angles at the intersection of parallel straight lines AC and BD secant AB.

From the similarity of triangles: AO / OB = CO / OD.

AO = AB – ОВ, and since according to the condition AB = СD, then AO = СD – ОВ.

CO = CD – OD.

Then: (СD – ОВ) / ОВ = (СD – ОD) / ОD.

ОВ * СD – ОВ * ОD = ОD * СD – ОВ * ОD.

ОВ * СD = ОD * СD.

ОВ = ОD, which means that the triangle BOD is isosceles, which was required to be proved.