Statement : If the perpendicular bisectors of two sides of a triangle meet at point that belongs to the third side then this is a right angle triangle.
Prove:
Let ABC is triangle,
In which DF and EF are the perpendicular bisectors of legs AB and BC respectively.
That is, ∠ FDB = ∠BEF= 90°
Where, [tex]F\in AC[/tex]
Then By the property of the circumcenter,
F must be mid point of leg AC.
That is, DF ║ BC
⇒ ∠ FDB + ∠ DBC = 180° ( the sum of two adjacent angles on the parallel lines by the same transversal is supplementary )
But, ∠ FDB = 90°
⇒∠ DBC = 90°
⇒ ∠ ABC = 90°
That is, Δ ABC is a right angle triangle.
