2007 Indonesia MO Problems/Problem 1

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Problem

Let $ABC$ be a triangle with $\angle ABC=\angle ACB=70^{\circ}$. Let point $D$ on side $BC$ such that $AD$ is the altitude, point $E$ on side $AB$ such that $\angle ACE=10^{\circ}$, and point $F$ is the intersection of $AD$ and $CE$. Prove that $CF=BC$.

Solution

[asy] pair a=(5,13.737),b=(0,0),c=(10,0),d=(5,0),e=(3.867,10.623),f=(5,8.660);  draw(a--b--c--a); draw(c--e); draw(a--d);  dot(a); label("$A$",a,N); dot(b); label("$B$",b,SW); dot(c); label("$C$",c,SE); dot(d); label("$D$",d,S); dot(e); label("$E$",e,NW); dot(f); label("$F$",f,SW);  draw(rightanglemark(c,d,a,40)); draw(anglemark(c,b,a,40)); label("$70^\circ$",anglemark(c,b,a),3*NE); [/asy]

By using the Angle Addition Postulate and substitution, \begin{align*} \angle ACB &= \angle ACE + \angle ECB \\ 70^\circ &= 10^\circ + \angle ECB \\ 60^\circ &= \angle ECB. \end{align*} Since $\triangle FDC$ is a 30-60-90 triangle, $FC = 2 \cdot CD$.


Additionally, because $D$ is the foot of the altitude from $A$, $AD \perp BC$. Furthermore, by the Base Angle Converse, we must have $AB = AC$ because $\angle ABC = \angle ACB$. Therefore, by AAS Similarity, $\triangle ADB \cong \triangle ADC$, so $BD = CD$. Thus, $BC = 2 \cdot CD$.


Therefore, by substitution, $BC = FC$.

See Also

2007 Indonesia MO (Problems)
Preceded by
First Problem
1 2 3 4 5 6 7 8 Followed by
Problem 2
All Indonesia MO Problems and Solutions
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