2007 Indonesia MO Problems/Problem 1

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, $BC = FC$ .

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

Page

Toolbox

Search

2007 Indonesia MO Problems/Problem 1

Problem

Solution

See Also