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,