Fie triunghiul dreptunghic $ABC$ cu ipotenuza $BC$ si $m(\angle ABC)=60^{\circ}$. Fie $M\in (BC)$ astfel incat $AB+BM=AC+CM$. Gasiti masura $\angle CAM$.
N. MIN HA
Triunghi 1 [ slicing ]
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Re: Triunghi 1 [ slicing ]
Considerăm punctul D pe latura AC astfel încât AB=AD și punctul E pe prelungirea laturii BC dincolo de C astfel încât CE=CD.Urmează că CE=CD=AC-AB=BM-MC.Deci BM=MC+CE,de unde M este mijlocul lui [BE].Ne folosim și de faptul că $m(\widehat{ABC})=60^\circ$.Obținem ușor că $m(\widehat{DBE})=m(\widehat{DEB})=15^\circ$.Astfel triunghiul BDE este isoscel cu [DM] mediană,deci $m(\widehat{DMB})=90^\circ$.Patrulaterul ADMB este inscriptibil,căci$m(\widehat{BAD})+m(\widehat{DMB})=180^\circ$.Folosind aceasta avem că$m(\widehat{DBM})=m(\widehat{DAM})=15^\circ$.În concluzie $m(\widehat{CAM})=15^\circ$.
- sunken rock
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Re: Triunghi 1 [ slicing ]
Another approach:
Let $I$ be the ıncenter of $\triangle ABC, N$ the contact point of incircle with $BC$. Well known: $M$ is the reflection of $N$ about $D$, midpoint of $BC$, so $DM=DN (1)$.
By symmetry, $\angle BDI=\angle BAI=45^\circ$, so $DN=NI (2)$. From $(2)$ we get $BN=CM (3)$ also.
$\triangle BIN\sim\triangle CBA$, so $\frac{IN}{BN}=\frac{AB}{AC} (4)$. With $(1),(2), (3)$ and $AB=AD$, from $(4)$ we get $\frac{DM}{CM}=\frac{AD}{AC}$, or $AM$ is the bisector of $\angle DAC=30^\circ$, done.
Best regards,
sunken rock
Let $I$ be the ıncenter of $\triangle ABC, N$ the contact point of incircle with $BC$. Well known: $M$ is the reflection of $N$ about $D$, midpoint of $BC$, so $DM=DN (1)$.
By symmetry, $\angle BDI=\angle BAI=45^\circ$, so $DN=NI (2)$. From $(2)$ we get $BN=CM (3)$ also.
$\triangle BIN\sim\triangle CBA$, so $\frac{IN}{BN}=\frac{AB}{AC} (4)$. With $(1),(2), (3)$ and $AB=AD$, from $(4)$ we get $\frac{DM}{CM}=\frac{AD}{AC}$, or $AM$ is the bisector of $\angle DAC=30^\circ$, done.
Best regards,
sunken rock
A blind man sees the details better.
- sunken rock
- Mesaje: 645
- Membru din: Joi Ian 06, 2011 2:49 pm
- Localitate: Constanta
Re: Triunghi 1 [ slicing ]
It's well known that $M$ is the contact point of the $A-$excircle of $\triangle ABC$ and, if $E,F$ are its contact points with $AB,AC$ respectively, then the circumcenter of $\triangle EFM$ is the reflection of $A$ in $EF$, with $AEOF$ a square, thus $EO=EA$ and, since $\angle EFM=30^\circ$, then $EF=EM$ and $\angle AEM=30^\circ$, that is, $\angle EAM=\angle AME=75^\circ$, making $\angle MAC=15^\circ$.
Best regards,
sunken rock
Best regards,
sunken rock
A blind man sees the details better.