# Ceva I.2

## Problem

• Let $M$ be the midpoint of side $AB$ of triangle $ABC$. Points $D$ and $E$ lie on line segments $BC$ and $CA$, respectively, such that $DE$ and $AB$ are parallel. Point $P$ lies on line segment $AM$. Lines $EM$ and $CP$ intersect at $X$ and lines $DP$ and $CM$ meet at $Y$. Prove that $X,Y,B$ are collinear.

## Solution

$[asy] import olympiad; size(12cm); pair A=origin, B=(12,0), C=(4,8); draw(A--B--C--cycle); dot("A",A,S); dot("B",B,S); dot("C",C,N); pair D=(7,5), E=(2.5,5); dot("D",D,NE); dot("E",E,NW); draw(D--E); path p = A--B; pair M=midpoint(p); dot("M",M,S); pair P=(4.5,0); dot("P",P,S); path x = E--M; path y = C--P; draw(E--M, yellow+linewidth(1)); draw(C--P, yellow+linewidth(1)); pair[] i = intersectionpoints(x,y); dot("X",i[0],W); path g = D--P; path h = C--M; draw(D--P, purple+linewidth(1)); draw(C--M, purple+linewidth(1)); pair[] j = intersectionpoints(g,h); dot("Y",j[0],W-dir(20)); draw(i[0]--j[0]--B, black+dashed+linewidth(0.5)); pair G=E+3.7*dir(125); dot("G",G,N); draw(E--G, black+dashed+linewidth(0.5)); draw(C--G, black+dashed+linewidth(0.5)); [/asy]$

We want to prove $X,Y,B$ collinear, so we consider from which which direction we want to prove this. We can prove $\angle XYC + \angle BYC = 180$ to do it, but we don't know any angles, so that probably won't be much use. Instead, we can prove $CM, DP, XB$ collinear, since the intersection of $CM$ and $DP$ is $Y$. So, let's consider Ceva's Theorem (a concurrency related formula) on $\triangle BCP$.

Let $AB = c, AC = b, BC = a$. That means $AM = MB = \frac{c}{2}$. There are a lot of unknowns here, so let further set $AP = x, CD = y$. We know that $$\dfrac{PM}{MB} \cdot \dfrac{BD}{DC} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{CX}{XP}$$ Now, if we extend $EM$ through $E$ and intersect the line at $C$ parallel to $AB$ at point $G$, we see $\triangle GEC \sim \triangle MEA$. Thus, $\dfrac{GC}{AM} = \dfrac{EC}{AE} = \dfrac{CD}{BD} \implies \dfrac{GC}{\frac{c}{2}} = \dfrac{y}{a-y} \implies GC = \dfrac{\frac{c}{2} \cdot y}{a-y}$. Using $\triangle GCX \sim \triangle MPX$, $\dfrac{CX}{XP} = \dfrac{GC}{PM} = \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}$. Thus, $$\dfrac{PM}{MB} \cdot \dfrac{BD}{DC} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}=1$$

$\mathcal{QED}$