Draw two isosceles triangles, ∆ABC and ∆ADC with common base AC . Vertexes B and D are in the opposite semi-planes determined by AC . Draw the line segment BD . Prove: AC ⊥ BD
Segment AC is a common base to ∆ABC and ∆ADC Given
Let K be the midpoint of the base of segment AC, then BK is the perpendicular bisector of segment perpendicular bisector of isosceles AC. triangle property.
Also, DK is the perpendicular bisector of segment AC. perpendicular bisector of isosceles triangle property.