[Toán 7] bài tập

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ngolong898

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soicon_boy_9x

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Ta có: B^=C^=180o40o2=70o\hat{B}=\hat{C}=\dfrac{180^o-40^o}{2}=70^o

$\rightarrow \widehat{BFI}=\widehat{BFA}+\widehat{AFI}=\hat{C}+
\widehat{FBC}+60^o=70^o+30^o+60^o=160^o$

Ta có: EBF^=70o30o30o=10o(1)\widehat{EBF}=70^o-30^o-30^o=10^o(1)

Ta lại có: FBA^=B^FBC^=70o30o=40o\widehat{FBA}=\hat{B}-\widehat{FBC}=70^o-30^o=40^o

ΔAFB\rightarrow \Delta AFB cân tại F

BF=AF\rightarrow BF=AFAF=FIAF=FI

BF=FIΔBFI\rightarrow BF=FI \rightarrow \Delta BFI cân tại F

IBF^=180o160o2=10o(2)\rightarrow \widehat{IBF}=\dfrac{180^o-160^o}{2}=10^o(2)

Từ (1)(1)(2)dpcm(2) \rightarrow dpcm


 
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