Congruent triangles





Let’s prove one direction: (1). Consider
shown below:
We showed previously that in any triangle with orthocenter
. Thus
. By one of the equivalent statements here, we conclude that equation (1) holds.





Similar to example 1 above. If is the reflection of the circumcenter over side
, then we always have
in any triangle. This would then lead to
, which by the equivalent statements here implies equation (1).









This follows from the two preceding examples, together with the fact that the segment has length
. (Recall that
is congruent to the parent
in such a way that
,
, and
.)
Cyclic trapezium




Note that under the assumption of equation (1), we have that the reflection lies on side
. Let
be the circumradius of the parent triangle
. Then
and also
. Since radius
is parallel to side
(see here), we get an isosceles trapezium
.








Similar to example 4 above.
Takeaway
In , let
be the side-lengths,
the circumcenter,
the reflection of
over side
,
the reflection of
over side
, and
the reflection of
over side
. Then the following statements are equivalent:
As usual, we have more of these in the exercises.
Task
- (Mid fifties) In a non-right triangle
, let
be the side-lengths,
the altitudes,
the feet of the altitudes from the respective vertices,
the circumradius,
the circumcenter,
the nine-point center,
the orthocenter,
the midpoint of side
,
the reflection of
over side
,
the reflection of
over side
, and
the reflection of
over side
. PROVE that the following fifty-six statements are equivalent:
is congruent to
is isosceles with
is isosceles with
is right angled at
is the circumcenter of
is right-angled at
is right-angled at
- quadrilateral
is a rectangle
- the points
are concyclic with
as diameter
- the reflection
of
over
lies internally on
- the reflection
of
over
lies externally on
- radius
is parallel to side
is the reflection of
over side
- the nine-point center lies on
- the orthic triangle is isosceles with
- the geometric mean theorem holds
- the bisector of
has length
, where
- the orthocenter is a reflection of vertex
over side
- segment
is tangent to the circumcircle at point
- median
has the same length as the segment
- the bisector
of
is tangent to the nine-point circle at
is a convex kite with diagonals
and
- altitude
is tangent to the nine-point circle at
- segment
is tangent to the nine-point circle at
.
(short of the target.)
- (Extra feature) If
satisfies equation (??), PROVE that its nine-point center
divides
in the ratio
.