Prove that the perpendicular bisector of EF also bisects BC.
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You know of course that the following statements:
(1) Perpendicular bisector of a chord in a circle is a circle diameter.
(2) The circumcenter of a right triangle is the midpoint of the hypotenuse.
Look at the following picture from my answer to a similar question some time ago:
http://farm6.static.flickr.com/5085/5325…
(on the picture A', B' and C' are the feet of the altitudes instead of D, E and F)
Triangles BB'C and CC'B are right, so their common circumcenter is the midpoint of BC. Both triangles have the same circumcircle, the latter means that the quadrilateral BCB'C' is cyclic (and by the way, triangles ABC and AB'C' are similar, etc.). Now according (1) above the perpendicular bisector of the chord B'C' (Your EF) passes through the center of the circumscribed circle of BCB'C' - the midpoint of BC.
(1) Perpendicular bisector of a chord in a circle is a circle diameter.
(2) The circumcenter of a right triangle is the midpoint of the hypotenuse.
Look at the following picture from my answer to a similar question some time ago:
http://farm6.static.flickr.com/5085/5325…
(on the picture A', B' and C' are the feet of the altitudes instead of D, E and F)
Triangles BB'C and CC'B are right, so their common circumcenter is the midpoint of BC. Both triangles have the same circumcircle, the latter means that the quadrilateral BCB'C' is cyclic (and by the way, triangles ABC and AB'C' are similar, etc.). Now according (1) above the perpendicular bisector of the chord B'C' (Your EF) passes through the center of the circumscribed circle of BCB'C' - the midpoint of BC.