Triunghi ECHILATERAL, o bisectoare si o paralela la o latura

mihai miculita
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Triunghi ECHILATERAL, o bisectoare si o paralela la o latura

Mesaj de mihai miculita »

Fie $ABC-$ un triunghi echilateral si $D$ un punct arbitrar al laturii $[AC].$ Notam cu $E-$punctul de intersectie al bisectoarei unghiului $\widehat{ABD},$ cu paralela dusa prin varful $A$ la latura $[BC].$ Aratati ca are loc relatia: $\boxed{|BD|=|AE|+|CD|}.$
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Andi Brojbeanu
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Re: Triunghi ECHILATERAL, o bisectoare si o paralela la o la

Mesaj de Andi Brojbeanu »

Daca $AB=BC=CA=a$ si $CD=x$, atunci $BD^2=BC^2+DC^2-2BC\cdot DC\cdot \cos{60^{\circ}}=a^2+x^2-ax$.
De asemenea, din asemanarea triunghiurilor $AEF$ si $CBF$ rezulta
$\displaystyle \frac{AE}{BC}=\frac{AF}{CF}=\frac{AF}{a-AF}=\frac{1}{\frac{a}{AF}-1}=\frac{1}{\frac{a}{\frac{AB\cdot AD}{BA+BD}}-1}=\frac{1}{\frac{a+BD}{a-x}-1}$$\displaystyle=\frac{a-x}{a+BD-a+x}=\frac{a-x}{BD+x}\Rightarrow AE=\frac{a(a-x)}{BD+x}$.
Atunci, $AE+CD=\displaystyle \frac{a(a-x)}{BD+x}+x=\frac{a^2-ax+x^2+xBD}{BD+x}=\frac{BD^2+xBD}{BD+x}$$=BD\Rightarrow BD=AE+CD$ q.e.d.
Brojbeanu Andi Gabriel, clasa XI-a
Colegiul National "Constantin Carabella" Targoviste
mihai miculita
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Re: Triunghi ECHILATERAL, o bisectoare si o paralela la o la

Mesaj de mihai miculita »

Eu aveam urmatoarea solutie trigonometrica:
Notand cu $u=m(\widehat{ABE})=m(\videhat{EBD}),$ si folosind teorema sinusurilor, din:
$\Delta{ABE}\Rightarrow\dfrac{|AB|}{\sin(60^0-u)}=\dfrac{|AE|}{\sin u}$$\Rightarrow|AB|=\dfrac{|AE|.\sin(60^0-u)}{\sin u};\,\,(1)$
$\Delta{ABD}\Rightarrow\dfrac{|AB|}{\sin(120^0-u)}=\dfrac{|BD|}{\sin 60^0}$$\Rightarrow|AB|=\dfrac{|BD|.\sin(120^0-2u)}{\sin 60^0}=$$\dfrac{2.|BD|.\sin(60^0-u).\cos(60^0-u)}{\sin 60^0};\,\,(2)$
$\Delta{BCD}\Rightarrow\boxed{\dfrac{|CD|}{|BD|}=\dfrac{\sin(60^0-2u)}{\sin 60^0}}.\,\,(3)$
Iar din relatiile (1) si (2)$\Rightarrow \dfrac{|AE|.\sin(60^0-u)}{\sin u}=\dfrac{2.|BD|.\sin(60^0-u).\cos(60^0-u)}{\sin 60^0}\Rightarrow$
$\Rightarrow\boxed{\dfrac{|AE|}{|BD|}=\dfrac{2.\sin u.\cos(60^0-u)}{\sin 60^0}}.\,\,(4)$
In fine, adunand acum relatiile (3) si (4), membru cu membru, obtinem:

$\dfrac{|AE|+|CD|}{|BD|}=\dfrac{2.\sin u.\cos(60^0-u)+\sin(60^0-2u)}{\sin 60^0}=$
$=\dfrac{[\sin 60^0-\sin(60^0-2u)]+\sin(60^0-2u)}{\sin 60^0}=\dfrac{\sin 60^0}{\sin 60^0}=1\Rightarrow$
$\Rightarrow\dfrac{|AE|+|CD|}{|BD|}=1\Rightarrow \boxed{|BD|=|AE|+|CD|}.$
Virgil Nicula
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Re: Triunghi ECHILATERAL, o bisectoare si o paralela la o la

Mesaj de Virgil Nicula »

Vedeti problema propusa PP6 aici.
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sunken rock
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Re: Triunghi ECHILATERAL, o bisectoare si o paralela la o la

Mesaj de sunken rock »

Rotate triangle BCD about B to send C to A and D to P on AE. Triangle BDP is equilateral, AP=CD ( 1 ), and BE is also bisector of angle PBC, hence triangle PBE is isosceles, thus PE=PB=BD; with (1) we are done.

Remarks: 1) Angle BED=30 degs.
2) The proof also shows that BD=CD+AE iff BE is angle bisector of angle ABD, when ABC is an equilateral triangle

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sunken rock
A blind man sees the details better.
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