Chứng minh tam giác MNP đều biết tam giác ABC đều, BD=1/3BC, AE=1/3AB

\(\Delta\)ABC đều =>AB=BC=CA và \(\widehat{ABC}=\widehat{BAC}=\widehat{ACB}\)

Xét \(\Delta\)ACE và \(\Delta\)BAD có:

AC=AB

\(\widehat{BAC}=\widehat{ABC}\)

AE=BD(=\(\dfrac{1}{3}\) độ dài cạnh \(\Delta\)ABC)

=>\(\Delta\)ACE=\(\Delta\)BAD(c.g.c)

=>\(\widehat{C_1}=\widehat{A_1}\)

Chứng minh tương tự ta có:\(\widehat{A_1}=\widehat{B_1}\)

=>\(\widehat{A_1}=\widehat{B_1}\)\(=\widehat{C_1}\)(1)

Mà \(\widehat{A_1}+\widehat{A_2}=\widehat{B_1}+\widehat{B_2}=\widehat{C_1}+\widehat{C_2}\left(=60^o\right)\)

=>\(\widehat{A_2}=\widehat{B_2}=\widehat{C_2}\)(2)

Lại có:\(\widehat{F_1}=\widehat{B_1}+\widehat{C_2}\)(t/c góc ngoài)(3)

\(\widehat{N_1}=\widehat{B_2}+\widehat{A_1}\)(t/c góc ngoài)(4)

\(\widehat{M_1}=\widehat{A_2}+\widehat{C_1}\)(t/c góc ngoài)(5)

Từ (1);(2);(3);(4) và (5)=>\(\widehat{M_1}=\widehat{N_1}=\widehat{P}_1\)

Mà: \(\widehat{M_1}=\widehat{M_2};\widehat{N_1}=\widehat{N_2};\widehat{P_1}=\widehat{P_2}\)(các góc đối đỉnh)

=>\(\widehat{M_2}=\widehat{N_2}=\widehat{P}_2\)

=>\(\Delta MNP\)đều(đpcm)