I was wondering if someone could show me how to do prove these in a Euclidean way:
I need to show that the following three statements are equivalent for the triangle ABC,
gamma (the angle opposite AB) = 90 degrees
C is the orthocenter of ABC
the midpoint of AB is the circumcenter of ABC
I think I can prove this if I show the first implies the second, the second implies the third, and the third implies the first. I drew out diagrams and I see that it makes sense, but I'm having trouble writing it out in a more rigorous way.
Any help would be appreciated. Thanks in advance.
(1) => (2) since , then AC (BC) is the altitude w/ respect to vertex A (B) [which both contain C]. The altitude with respect to C passes also C. The three altitudes intersect at C, therefore, C is the orthocenter of .
Originally Posted by PvtBillPilgrim
(2) => (3) i'll try to work on this tomorrow.
(3) => (1) Let D, the midpoint of AB, be the circumcenter of ABC. Let E be on CB such that DE is the perpendicular bisector of CB, and let F be on CA such that DF is the perpendicular bisector of CA. Then, CFDE forms a quadrilateral. In particular, since angles CDF and DEC forms right angles, then CFDE is a rectangle. therefore, angle C is a right angle.
(2) => (3) Let C be the orthocenter of ABC. then the altitude from A to BC is AC which means AC forms a right angle with BC. Let be the midpoint of AB and consider a point outside the triangle, say D, such that ADBC forms a rectangle. Then, AB is a diagonal of the rectangle.
Note that the diagonals of a rectangle bisect each other. Thus, is in the diagonal CD. Therefore is congruent to .
Let be on CB such that perpendicular to CB.
now, ; and is congruent to . Therefore, by HL theorem, (noting that the triangles are right.). By CPCTC Theorem, ; therefore, is the perpendicular bisector of CB.
similarly, if be on CA such that perpendicular to CA, it can be shown that is the perpendicular bisector of CA.
The perpendicular bisector of AB passes . Since all the perpendicular bisectors of each side intersect at , therefore is the orthocenter of ABC..
PS: drawing will help you in the visualization of the proof..