Ceva I.2
Problem
- Let
be the midpoint of side
of triangle
. Points
and
lie on line segments
and
, respectively, such that
and
are parallel. Point
lies on line segment
. Lines
and
intersect at
and lines
and
meet at
. Prove that
are collinear.
Solution
[asy] import olympiad; size(12cm);
pair A=origin, B=(12,0), C=(4,8);
draw(A--B--C--cycle);
dot("",A,S); dot("
",B,S); dot("
",C,N);
pair D=(7,5), E=(2.5,5);
dot("",D,NE); dot("
",E,NW);
draw(D--E);
path p = A--B;
pair M=midpoint(p); dot("
",M,S);
pair P=(4.5,0); dot("",P,S);
path x = E--M; path y = C--P;
draw(E--M, yellow+linewidth(1)); draw(C--P, yellow+linewidth(1));
pair[] i = intersectionpoints(x,y); dot("
",i[0],W);
path g = D--P; path h = C--M;
draw(D--P, purple+linewidth(1)); draw(C--M, purple+linewidth(1));
pair[] j = intersectionpoints(g,h); dot("
",j[0],W-dir(20));
draw(i[0]--j[0]--B, black+dashed+linewidth(0.5));
pair G=E+3.7*dir(125); dot("",G,N);
draw(E--G, black+dashed+linewidth(0.5)); draw(C--G, black+dashed+linewidth(0.5));
[/asy]
We want to prove collinear, so we consider from which which direction we want to prove this. We can prove
to do it, but we don't know any angles, so that probably won't be much use. Instead, we can prove
collinear, since the intersection of
and
is
. So, let's consider Ceva's (a concurrency related formula) on
.
Let . That means
. There are a lot of unknowns here, so let further set
. We know that
Now, if we extend
through
and intersect the line at
parallel to
at point
, we see
. Thus,
. Using
,
. Thus,