Problema Canada Olymon 2002

[radu alberto]

$\triangle{ABC}$ is a triangle with $\angle{A}$$=$$40^\circ$ and $\angle{B}$$=$$60^\circ$.Let $D$ and $E$ be respective points of $AB$ and $AC$ for which $\angle{DCB}$$=$$70^\circ$ and $\angle{EBC}$$=$$40^\circ$. Furthermore, let $F$ be the point of intersection of $DC$ and $EB$. Prove that $AF$ and $BC$ are perpendiculare.

[Calin Spiridon]

O solutie sintetica:
Se vede usor ca $m(\widehat{BCF})=m(\widehat{CFB})=70^\circ$,de unde BF=BC,foarte util pe viitor.Sa spunem ca AF taie pe BC in H.Trebuie aratat ca $\widehat{AHB}$ este drept.Atunci ar trebui demonstrat ca $m(\widehat{HAC})=10^\circ{$.Aceeasi masura o are si $\widehat{FCA}$,ceea ce ne sugereaza ca pentru a rezolva problema aratam ca AF=FC.Vom face o constructie auxiliara,care sa ne duca la o banala congruenta de triunghiuri.Construim pe latura BC,in exterior,triunghiul echilateral CBG.Avem BF=BC=BG=GC.BGF-isoscel cu unghiul din varf de $100^\circ$,de unde $m(\widehat{BGF})=40^\circ, m(\widehat{FGC})=20^\circ$.De asemenea avem $m(\widehat{GCF})=m(\widehat{BFA})=130^\circ$De aici triunghiurile BFA si GCF sunt congruente,cazul ULU.Astfel,AF=FC,q.e.d.

[sunken rock]

O solutie sintetica:
Se vede usor ca $m(\widehat{BCF})=m(\widehat{CFB})=70^\circ$,de unde BF=BC,foarte util pe viitor.Sa spunem ca AF taie pe BC in H.Trebuie aratat ca $\widehat{AHB}$ este drept.Atunci ar trebui demonstrat ca $m(\widehat{HAC})=10^\circ{$.Aceeasi masura o are si $\widehat{FCA}$,ceea ce ne sugereaza ca pentru a rezolva problema aratam ca AF=FC.Vom face o constructie auxiliara,care sa ne duca la o banala congruenta de triunghiuri.Construim pe latura BC,in exterior,triunghiul echilateral CBG.Avem BF=BC=BG=GC.BGF-isoscel cu unghiul din varf de $100^\circ$,de unde $m(\widehat{BGF})=40^\circ, m(\widehat{FGC})=20^\circ$.De asemenea avem $m(\widehat{GCF})=m(\widehat{BFA})=130^\circ$De aici triunghiurile BFA si GCF sunt congruente,cazul ULU.Astfel,AF=FC,q.e.d.

How did you get $\angle BFA=130^\circ$ ???

Take also $K\in AC|BK=BC$, see that $\Delta BCK$ is isosceles with $\angle BKC=80^\circ\implies\Delta ABK$ isosceles with angles $40^\circ-40^\circ-100^\circ$, hence$\Delta ABK\sim\Delta GFB$; having $BK=BC=BG$ we infer they are actually congruent, hence $FG=AB$. Now, indeed $\Delta GCF\cong\Delta BFA$ (s.a.s. criterion), consequently $AF=CF$ and we are done.


Best regards,
sunken rock

[Calin Spiridon]

Intr-adevar,greseala mea...,mersi oricum ca ai continuat rezolvarea incheiata brusc de mine din neatentie.

[sunken rock]

Another solution, maybe very nice!
As proven above, $BF=BC$. Let $M$ be the reflection of $F$ about $AB$ and $P=AC\cap BM$. $\triangle BCP$ is a Langley triangle and, with $\angle PBA=20^\circ,\angle PCM=30^\circ$, we infer $\angle BAM=30^\circ$ and thus $\triangle AFM$ is equilateral, hence $AF=FM=CF$ and easily now $AF\bot BC$.

Best regards,
sunken rock