1. ## geometry

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.

2. Originally Posted by PvtBillPilgrim
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,
(1) gamma (the angle opposite AB) = 90 degrees
(2) C is the orthocenter of ABC
(3) 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 $\displaystyle \angle C=\gamma=90^0$, 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 $\displaystyle \Delta ABC$.

(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.

3. (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 $\displaystyle M_{AB}$ 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, $\displaystyle CM_{AB}$ is in the diagonal CD. Therefore $\displaystyle CM_{AB}$ is congruent to $\displaystyle M_{AB}B$.
Let $\displaystyle M_{CB}$ be on CB such that $\displaystyle M_{AB}M_{CB}$ perpendicular to CB.

now, $\displaystyle M_{AB}M_{CB}=M_{AB}M_{CB}$; and $\displaystyle CM_{AB}$ is congruent to $\displaystyle M_{AB}B$. Therefore, by HL theorem, $\displaystyle \Delta CM_{CB}M_{AB} \cong \Delta BM_{CB}M_{AB}$ (noting that the triangles are right.). By CPCTC Theorem, $\displaystyle CM_{CB} \cong BM_{CB}$; therefore, $\displaystyle M_{AB}M_{CB}$ is the perpendicular bisector of CB.

similarly, if $\displaystyle M_{CA}$ be on CA such that $\displaystyle M_{AB}M_{CA}$ perpendicular to CA, it can be shown that $\displaystyle M_{AB}M_{CA}$ is the perpendicular bisector of CA.

The perpendicular bisector of AB passes $\displaystyle M_{AB}$. Since all the perpendicular bisectors of each side intersect at $\displaystyle M_{AB}$, therefore $\displaystyle M_{AB}$ is the orthocenter of ABC..