Apparently I haven't forgotten as much synthetic geo as expected
by math_explorer, May 19, 2014, 2:16 PM
I was procrastinating and discovered a file called "Is it obvious.pdf" in my computer. Somehow, I remembered where it came from: http://www.artofproblemsolving.com/Forum/blog.php?u=56957&b=82127
I then solved it again (without peeking at my old solution) (and without using any scratch paper) (or any dynamic geo software) to avoid doing history homework, and like my new solution better, although I guess the essence is the same.
Whatever, /post in May
Let
and 
denote the circumcircles of 
and 
, respectively (whose centers are respectively 
and 
, as you'd expect.)
Note that since the two triangles are congruent, their circumcircles are congruent as well. Then, since
is the midpoint of 
, the circles and are reflections about .
Reflect
and 
about to get 
and 
. Since and are on , we know that and are on . Also, since the triangles are congruent, the arc subtended (?) by 
on equals the arc subtended by 
on equals the arc subtended by 
on . This arc equality shows 
.
Now note that since
is a medial segment of 
and 
is a medial segment of 
, we have 
and 
are collinear.
I then solved it again (without peeking at my old solution) (and without using any scratch paper) (or any dynamic geo software) to avoid doing history homework, and like my new solution better, although I guess the essence is the same.
Whatever, /post in May
Let
and 
denote the circumcircles of 
and 
, respectively (whose centers are respectively 
and 
, as you'd expect.)Note that since the two triangles are congruent, their circumcircles are congruent as well. Then, since
is the midpoint of 
, the circles and are reflections about .Reflect
and 
about to get 
and 
. Since and are on , we know that and are on . Also, since the triangles are congruent, the arc subtended (?) by 
on equals the arc subtended by 
on equals the arc subtended by 
on . This arc equality shows 
.Now note that since
is a medial segment of 
and 
is a medial segment of 
, we have 
and 
are collinear.This post has been edited 1 time. Last edited by math_explorer, May 19, 2014, 2:18 PM
and 
cut the sides at their midpoints 
and are congruent, hence 
and we want to show the midpoints of 
,
an 
are collinear. But by Newton's line, it suffices to show the midpoints of 
and 
are collinear, where 
and 
.
(why?), which is true as both are equal to 
by similar triangles.
May 2020
January 2020