Question

bài 2: cho \(\Delta ABC\) , \(\widehat{B}=60^o\) . hai đường phân giác AP và CQ cắt nhau tại I.CM

a) \(\widehat{AIC}=?\)

b) IQ = IP

Answer 1

Bài 2:

Áp dụng tc tổng 3 góc trong 1 tg ta có:

\(\widehat{ABC}\) + \(\widehat{BAC}\) + \(\widehat{BCA}\) = 180^o

=> \(\widehat{BAC}\) + \(\widehat{BCA}\) = 180^o - \(\widehat{ABC}\)

=> \(\widehat{BAC}\) + \(\widehat{BCA}\) = 180^o - 60^o

=> \(\widehat{BAC}\) + \(\widehat{BCA}\) = 120^o

Ta có: \(\widehat{IAC}\) = \(\frac{1}{2}\) \(\widehat{BAC}\) (AI là tia pg)

\(\widehat{ICA}\) = \(\frac{1}{2}\) \(\widehat{BCA}\) (CI là tia pg)

=> \(\widehat{IAC}\) + \(\widehat{ICA}\) = \(\frac{1}{2}\) \(\widehat{BAC}\) + \(\frac{1}{2}\) \(\widehat{BCA}\)

=> \(\widehat{IAC}\) + \(\widehat{ICA}\) = \(\frac{1}{2}\) (\(\widehat{BAC}\) + \(\widehat{BCA}\))

=> \(\widehat{IAC}\) + \(\widehat{ICA}\) = \(\frac{1}{2}\). 120^o = 60^o

Áp dụng tc tổng 3 góc trong 1 tg ta có:

\(\widehat{IAC}\) + \(\widehat{ICA}\) + \(\widehat{AIC}\) = 180^o

=> \(\widehat{AIC}\) = 180^o - ( \(\widehat{IAC}\) + \(\widehat{ICA}\))

=> \(\widehat{AIC}\) = 180^o - 60^o = 120^o

b) Nối B với I

Kẻ IE \(\perp\) BC; IH \(\perp\) AB và ID \(\perp\) AC

Ta có: \(\widehat{AIC}\) = \(\widehat{QIP}\) = 120^o (đối đỉnh)

Áp dụng tc tgv ta có:

\(\widehat{BIH}\) + \(\widehat{HBI}\) = 90^o

\(\widehat{BIE}\) + \(\widehat{IBE}\) = 90^o

=> \(\widehat{BIH}\) + \(\widehat{HBI}\) + \(\widehat{BIE}\) + \(\widehat{IBE}\) = 180^o

=> (\(\widehat{HBI}\) + \(\widehat{IBE}\)) + (\(\widehat{BIH}\) + \(\widehat{BIE}\)) = 180^o

=> \(\widehat{ABC}\) + (\(\widehat{BIH}\) + \(\widehat{BIE}\)) = 180^o

=> 60^o + \(\widehat{HIE}\) = 180

=> \(\widehat{HIE}\) = 120^o

=> \(\widehat{QIP}\) = \(\widehat{HIE}\)

Lại có: \(\widehat{QIE}\) + \(\widehat{EIP}\) = \(\widehat{QIP}\)

\(\widehat{QIE}\) + \(\widehat{QIH}\) = \(\widehat{HIE}\) mà \(\widehat{QIP}\) = \(\widehat{HIE}\) => \(\widehat{EIP}\) = \(\widehat{QIH}\) Xét \(\Delta\)HIA vuông tại H và \(\Delta\)DIA vuông tại D có: IA chung \(\widehat{HAI}\) = \(\widehat{DAI}\) (tia pg) => \(\Delta\)HIA = \(\Delta\)DIA (ch - gn) => HI = DI (2 cạnh t/ư) (1) Tương tự: \(\Delta\)EIC = \(\Delta\)DIC (ch - gn) => EI = DI (2 cạnh t/ư) (2) Từ (1) và (2) suy ra HI = EI. Xét \(\Delta\)QIH vuông tại H và \(\Delta\)PIE vuông tại E có: HI = IE (c/m trên) \(\widehat{EIP}\) = \(\widehat{QIH}\) (c/m trên) => \(\Delta\)QIH = \(\Delta\)PIE (ch - gn) => QI = PI (2 cạnh t/ư)