# 2007 IMO Problems/Problem 2

## Problem

Consider five points $A,B,C,D$, and $E$ such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral. Let $\ell$ be a line passing through $A$. Suppose that $\ell$ intersects the interior of the segment $DC$ at $F$ and intersects line $BC$ at $G$. Suppose also that $EF=EG=EC$. Prove that $\ell$ is the bisector of $\angle DAB$.

## Solution

$[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); path circle = Circle(origin, 1); draw(circle); pair D = (-sqrt(3)/2, -0.5), C = (sqrt(3)/2, -0.5); //G = bisectorpoint(C, B, D); pair Ee = rotate(38,C)*D; pair E = IP(C--Ee, circle,1); pair Gg = rotate(76,C)*D; path circle2 = Circle(E, length(C-E)); pair G = IP(C--Gg, circle2, 1); pair F = IP(C--D, circle2, 1); pair Bb = rotate(-104,C)*D; pair B = IP(C--Bb, circle, 1); pair A = extension((-1,B.y),(1,B.y),G,F); draw(circle2, dashed); draw(A--G); draw(C--D--A--B); draw(G--B); draw(E--F); draw(E--C); draw(E--G); dot("C", C, dir(30)); dot("D", D, SW); dot("G", G, SE); dot("E", E, SW); dot("F", F, SW); dot("A", A, SW); dot("B", B, SE); draw(D--E,dashed); draw(B--E,dashed); [/asy]$

Since $\angle{DAF}=\angle{CGF}$, $\angle{BAF}=\angle{CFG}$, it suffices to prove $CF=CG$.

Let $\angle{FCE}=\alpha$, $\angle{GCE}=\beta$, $\angle{CDE}=\gamma$. We have: $$CF=2CE\cos{\alpha}, CG=2CE\cos{\beta}$$ so, $$\dfrac{CF}{CG}=\dfrac{\cos{\alpha}}{\cos{\beta}}$$ Meantime, using Law of Sines on $\triangle{DEC}$, we have, $$\dfrac{CE}{\sin{\gamma}}=\dfrac{DC}{\sin{(180-\alpha-\gamma)}}=\dfrac{DC}{\sin{(\alpha+\gamma)}}$$ Using Law of Sines on $\triangle{BEG}$, and notice that $\angle{CBE}=\angle{CDE}=\gamma$, we have, $$\dfrac{CE}{\sin{\gamma}}=\dfrac{BG}{\sin{(180-\beta-\gamma)}}=\dfrac{BG}{\sin{(\beta+\gamma)}}$$ so, $$\dfrac{DC}{BG}=\dfrac{\sin{(\alpha+\gamma)}}{\sin{(\beta+\gamma)}}$$ Since $\triangle{GFC} \sim \triangle{GAB}$, and $DC=AB$, we have, $\dfrac{DC}{BG}=\dfrac{AB}{BG}=\dfrac{CF}{CG}$. Hence, $$\dfrac{\sin{(\alpha+\gamma)}}{\sin{(\beta+\gamma)}}=\dfrac{\cos{\alpha}}{\cos{\beta}}$$ or, $$\sin{(\alpha+\gamma)}\cos{\beta}=\sin{(\beta+\gamma)}\cos{\alpha}$$ $$\dfrac{1}{2}(\sin{(\alpha+\gamma+\beta)}-\sin{(\alpha+\gamma-\beta)})=\dfrac{1}{2}(\sin{(\alpha+\gamma+\beta)}-\sin{(\beta+\gamma-\alpha)})$$ $$\sin{(\alpha+\gamma-\beta)}=\sin{(\beta+\gamma-\alpha)}$$ There are two possibilities: (1) $\alpha+\gamma-\beta = \beta+\gamma-\alpha$, or (2) $\alpha+\gamma-\beta = 180 - (\beta+\gamma-\alpha)$. However, (2) would mean $\gamma=90$, then $EC$ would be a diameter, and $EF < EC$ because $F$ is inside the circle, so (2) is not valid. From condition (1), we have $\alpha=\beta$, therefore $CF=CG$. $\square$

Solution by Mathdummy

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

 2007 IMO (Problems) • Resources Preceded byProblem 1 1 • 2 • 3 • 4 • 5 • 6 Followed byProblem 3 All IMO Problems and Solutions