# 2002 Indonesia MO Problems/Problem 4

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Given a triangle $ABC$ with $AC > BC$. On the circumcircle of triangle $ABC$ there exists point $D$, which is the midpoint of arc $AB$ that contains $C$. Let $E$ be a point on $AC$ such that $DE$ is perpendicular to $AC$. Prove that $AE = EC + CB$.

## Solution $[asy] size(5cm); draw(unitcircle); pair A, C, B; A = (-0.6, 0.8); C = (0.6, 0.8); B = (0.8, 0.6); dot(A); dot(C); dot(B); label("A", A, NW); label("C", C, N); label("B", B, SE); draw(C--B--A); pair Bp; Bp = (0.88, 0.8); dot(Bp, blue); label("B'", Bp, NE, blue); draw(A -- Bp); pair midAB, dirAB; midAB = (A + B)/2; dirAB = (B - A); path perpAB; perpAB = (midAB - dir(90)*dirAB) -- (midAB + dir(90)*dirAB); draw(perpAB, blue); pair Ep, dirABp; path perpABp; Ep = (A + Bp)/2; dirABp = (Bp - A); perpABp = (Ep - dir(90)*dirABp) -- (Ep + dir(90)*dirABp); dot(Ep, blue); label("E'", Ep, NW, blue); draw(perpABp, blue); pair Dp; Dp = intersectionpoint(perpAB, perpABp); dot(Dp, blue); label("D'", Dp, NW, blue); draw(Bp--Dp--B, red); draw(C--Dp--A, red); [/asy]$

We use the method of phantom points.

Draw $AC$ and $BC$, and extend line $AC$ past $C$ to a point $B'$ such that $BC = B'C$. Draw point $E'$ at the midpoint of $AB'$, and $D'$ at the intersection of the perpendicular to $AC$ from $E'$ and the perpendicular bisector of $AB$.

Since $D'B = D'B', CB = CB', D'C = D'C$, we have $\triangle D'BC \cong \triangle D'B'C$ by side-side-side similarity. Then $\angle D'AC = \angle D'AE' = \angle D'B'E' = \angle D'B'C = \angle D'BC$, so $ADCB$ is cyclic.

In particular, since we have $AD' = B'D' = BD'$, we know that $D'$ must be the midpoint of the arc of the circumcircle of $\triangle ABC$ that contains point $C$, and since $D'$ was on the perpendicular to $AC$ from $E'$, we must have that $E'$ is the foot of the perpendicular of $D'$ to $AC$. But this uniquely identifies $D' = D, E' = E$, and we are done.