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 .

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.  

Draw sides AB and BC with the same length, but different as AC length, then, ∆ABC is isosceles. Draw ∆ADC analogously.

Using    side AC as common base, draw sides AD and BD with the same length, but different as AC length, then, ∆ADC is isosceles.

Locate the midpoint of AC, let's call it E. The segment EB must be perpendicular to side AC, otherwise sides AB and BC would not have the same length. For the same reason, segment ED is also perpendicular to side AC. Both EB and ED are perpendicular to AC and pass through the same point E, then the segment BD is formed, and is perpendicular to AC.

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step-by-step explanation:

c > -10.55

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