0
$\begingroup$

In the picture $\Delta ABC$ angle A equals 60 degrees, AD is a bisector of BC. DF is perpendecular to BC. The angle FDA equals the angle FDE. How can I prove that the point E coincides with the centre of circumscribed circle in the triangle ABC?

(In other words we have to prove that the centre of the circle belongs to DE)

enter image description here

  • 0
    $AD$is a bisector of what? – 2017-02-26
  • 0
    Of $\angle CAB$ – 2017-02-26
  • 0
    @student28 Are we given $\angle FDE$? – 2017-02-26
  • 0
    A bisector of angle BAC. – 2017-02-26
  • 0
    No, we are not. it is just to show equality. – 2017-02-26
  • 0
    Actually you can only prove that the center is on the line through $D$ and $E,$ not that it is specifically at $E.$ So rather than saying one thing which is not provable and then "in other words" a second thing, just say the second thing. – 2017-02-27

1 Answers 1

0

HINT.

Bisector $AD$ intersects arc $BC$ at its midpoint $H$, and so does the perpendicular bisector of side $BC$ (see diagram below, where $O$ is the circle center). The required angle equality is equivalent to proving that triangle $ODH$ is isosceles, which in turn is a consequence of $OCH$ being an equilateral triangle.

enter image description here