TRIÂNGULOS
Autor: Aloyana Couto da Silva | Publicado: 2013-03-03T11:10:00-08:00









Algumas Questões&nbsp;











&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Considere três pontos A, B e C, não
colineares, chama-se triângulo ABC à reunião (ΔABC = AB U AC U BC) dos
segmentos AB, AC e BC. Notado por ABC = ΔABC.



&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Duas figuras são congruentes se elas possuem a mesma forma e o mesmo tamanho.










&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;1) Teorema do Triângulo Isósceles:&nbsp;Se um triângulo é isósceles, os ângulos da base
são congruentes.









Hipótese&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Tese





&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (∆ABC, AB ≡ AC)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; →&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; B
≡ C





Dem.:



Dados
ABC e ACB triângulos. Associe a A, B e C, respectivamente, A, C e B.



Como
&nbsp;&nbsp;AB ≡ AC



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; BÂC ≡ CÂB



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; AC ≡ AB, sendo AC do triângulo ABC e
AB do triângulo ACB.



Então
∆ABC ≡ ∆ACB.



Portanto
B ≡ C&nbsp;&nbsp;&nbsp;&nbsp; ■



***



2)&nbsp;Se dois triângulos têm
ordenadamente congruentes um lado e os dois ângulos a ele adjacentes, então
esses triângulos são congruentes.











Hipótese&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Tese



(B ≡
B´ (I), BC ≡ B´C´ (II), C ≡ C´ (III)&nbsp;&nbsp;&nbsp;&nbsp;
→&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ∆ABC´ ≡ ∆A´B´C´



Dem.:



Pelo
Postulado do Transporte de Segmentos (Dados um segmento AB e uma
semi-reta de origem A´, existem sobre esta semi-reta um único ponto B´tal que
A´B´ seja congruentes a AB.), obtemos na semi-reta B´A´ um ponto X
tal que B´X ≡ BA (IV)



Assim,
(II) BC ≡ B´C´, (I) B ≡ B´, (IV) BA ≡ B´X.



Então
∆ABC ≡ ∆XB´C´&nbsp; → BCA ≡ B´C´X (V)



Da
hipótese (III) BCA ≡ B´CA´, com (V) BCA ≡ B´CX e com o Postulado do Transporte
de Ângulos (Dados um ângulo AOB e uma semi-reta O´A´de um plano,
existe sobre este plano, e num dos semiplanos que O´A´ permite determinar, uma
única semi-reta O´B´ que forma com O´A´ um ângulo A´O´B´congruente ao ângulo
AOB.), decorre que B´A´ e
C´X = C´A´ interceptam-se num único ponto X = A´.



De X
≡ A´, com (IV), vem que B´A´≡ BA.



Então



BA ≡
B´A´, B ≡ B´ e BC ≡ B´C´.



Logo,
∆ABC ≡ ∆A´B´C´&nbsp;&nbsp; ■



***





3) As medianas relativas aos
lados congruentes de um triângulo isósceles são congruentes.







&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Hipótese&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Tese



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;AB&nbsp; =
AC, &nbsp;BM e CN são medianas&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; →&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; BM = CN



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Dem.: &nbsp;&nbsp;

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Sejam
BAM e CAN triângulos.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Se&nbsp;&nbsp;&nbsp;&nbsp; BA = CA



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; A
= A



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; AM = NA



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Então
∆BAM ≡ ∆CAN



&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Logo,
BM = CN&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ■










***








4) Se dois triângulos
retângulos têm ordenadamente congruentes um cateto e a hipotenusa, então esses
triângulos são congruentes.









&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Hipótese &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;Tese



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; A ≡
A´(retos)(I), AB ≡ A´B´(II), BC ≡ B´C´(III)&nbsp;&nbsp;
→&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ∆ABC ≡ ∆A´B´C´



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Dem.:



Tome o ponto D na
semi-reta oposta à semi-reta A´C´ tal que A´D ≡ AC (Postulado do Transporte de
Segmento).



AB ≡ A´B´, A ≡ A´, AC ≡ A´D
daí ∆ABC ≡ ∆A´B´D então BC ≡ B´D(IV) e C ≡ D(V)



De (IV) e (III) vem B´C´ ≡
B´D então ∆B´C´D é isósceles de base C´D → C´≡ D (VI)



De (V) e (VI) → C ≡ C´



Considere os triângulos ABC e A´B´C´,
tem-se que:



BC ≡ B´C´, C ≡ C´, A ≡ A´.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Portanto, ∆ABC ≡ ∆A´B´C´&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ■&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;







***







Nota:&nbsp;



Todos os Triângulos







Isósceles não são congruentes



Equiláteros não são congruentes



Retângulos não são congruentes



Retângulos isósceles não são congruentes



Acutângulos não são congruentes





***













Para Saber Mais:



FACCNINI, Walter. Matemática
volume único, 2ª edição – 1997 – Editora Saraiva



DOLCE, Osvaldo. POMPEO, José
Nicolau. Fundamentos de Matemática Elementar, volume 9 – 1995 – Editora Atual.











Δ &nbsp;trilátero &nbsp;⍢