# 2007 Indonesia MO Problems/Problem 1

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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 byFirst Problem 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed byProblem 2 All Indonesia MO Problems and Solutions
Invalid username
Login to AoPS