Problema I. Pe laturile \( \triangle ABC \) construim in afara acestuia triunghiurile isoscele \( \triangle BMC \) , \( \triangle CNA \) , \( \triangle APB \)
asfel incat \( m(\widehat{BMC}) + m(\widehat{CNA}) + m(\widehat{APB})=360^{\circ} \) , adica \( m(\widehat {CBM}) + m(\widehat {ACN}) + m(\widehat {BAP}) = 90^{\circ} \) .
Sa se arate ca masurile unghiurilor triunghiului \( MNP \) nu depind de forma triunghiului si acestea sunt :
\( \left\|\begin{array}{c}
m(\widehat {PMN}) = 90^{\circ} - m(\widehat {CBM}) = \frac 12\cdot m(\widehat {BMC})\\\\
m(\widehat {MNP}) = 90^{\circ} - m(\widehat {ACN}) = \frac 12\cdot m(\widehat {CNA})\\\\
m(\widehat {NPM}) = 90^{\circ} - m(\widehat {BAP}) = \frac 12\cdot m(\widehat {APB})\end{array}\right\| \) .
Problema II. Pe laturtile \( \triangle ABC \) construim in afara acestuia \( \triangle ABF\ , \ \triangle ACE\ ,\ \triangle BCD \) astfel incat \( DB = DC \) ,
\( \left\|\begin{array}{c}
m(\widehat {BAF}) = m(\widehat {CAE}) = x\\\\
m(\widehat {ABF}) = m(\widehat {ACE}) = y\\\\
m(\widehat {BCD}) = m(\widehat {CBD})=z\end{array}\right\| \) si \( x + y + z = 90^{\circ} \) . Sa se arate ca \( DE = DF \) si \( m(\widehat {EDF}) = 2y \).
Caz particular remarcabil. \( z = 0\ (x + y = 90^{\circ}) \) , adica \( \left\|\begin{array}{c}
D\in (BC)\ ,\ DB = DC\\\\
FA\perp FB\ ,\ EA\perp EC\\\\
\widehat {BAF}\equiv \widehat {CAE}\end{array}\right\| \) .
Tema de cercetare. Gasiti concluzia fiecarei probleme in cazul cand (cel putin)
unul din cele trei triunghiuri este construit spre interiorul triunghiului dat..
Triunghiuri construite pe laturile unui triunghi dat
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PROBLEMA 1 Notam \( m(\angle CBM)=x, m(\angle ACN)=y,m(\angle ABP)=z \)
Aplicam teorema Pitagora generalizata in \( \triangle NAP \) Obtinem:
\( PN^2=AN^2+AP^2-2AN\cdot AP\cdot\cos\angle PAN \) Dar \( AN=\frac{b}{2\cos y}, AP=\frac{c}{2\cos z}, m(\angle PAN)=90+A-x \)
Dupa calcule se obtine ca \( 4NP^2\cos^2 y\cos^2 z=4S\cos x\cos y\cos z+a^2\sin x\cos y\cos z+b^2\sin y\cos x\cos z+c^2\sin z\cos x\cos y \) si analoagele.
Asadar\( \frac{NP}{\cos x}=\frac{MN}{\cos y}=\frac{MP}{\cos z} \) de unde rezulta ca triunghiul \( MNP \)are unghiurile \( 90-x,90-y,90-z \)
Aplicam teorema Pitagora generalizata in \( \triangle NAP \) Obtinem:
\( PN^2=AN^2+AP^2-2AN\cdot AP\cdot\cos\angle PAN \) Dar \( AN=\frac{b}{2\cos y}, AP=\frac{c}{2\cos z}, m(\angle PAN)=90+A-x \)
Dupa calcule se obtine ca \( 4NP^2\cos^2 y\cos^2 z=4S\cos x\cos y\cos z+a^2\sin x\cos y\cos z+b^2\sin y\cos x\cos z+c^2\sin z\cos x\cos y \) si analoagele.
Asadar\( \frac{NP}{\cos x}=\frac{MN}{\cos y}=\frac{MP}{\cos z} \) de unde rezulta ca triunghiul \( MNP \)are unghiurile \( 90-x,90-y,90-z \)