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

Accepted Solution

Let  ΔABC is an isosceles triangle. AB = BC and AC is non-equal side of the triangle. Let draw the median of side AC , which meets at vertex B  and bisect ∠B and bisect AC . We can mark mid point of AC as 'O'. So AC⊥BO. Similarly  ΔADC is an isosceles  triangle. Median of DC meets and bisect ∠D.The mid point of ADC of side AC is O. Then AC⊥ DO. When we join B and D It passes through mid point of AC. Therefore AC⊥BD.