Chứng minh IQ=IP biết tam giác ABC có góc B=60 độ, 2 đường phân giác AP, CQ

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