Đề kiểm tra 15 phút - Đề số 4 - Bài 3 - Chương 2 - Hình học 8

a) \(\dfrac{{{S_{HBC}}}}{{{S_{ABC}}}} = \dfrac{{HA'}}{{AA'}}\)

b) \(\dfrac{{HA'}}{{AA'}} + \dfrac{{HB'}}{{BB'}} + \dfrac{{HC'}}{{CC'}} = 1.\)

Lời giải

 

a) Ta có: \({S_{HBC}} = \dfrac{1}{2}BC.HA';\) \({S_{ABC}} = \dfrac{1}{2}BC.AA'\)

\( \Rightarrow \dfrac{{{S_{HBC}}}}{{{S_{ABC}}}} = \dfrac{{HA'}}{{AA'}}\)

b) Chứng minh tương tự câu a) ta có:

\(\dfrac{{{S_{HAC}}}}{{{S_{ABC}}}} = \dfrac{{HB'}}{{BB'}}\) và \(\dfrac{{{S_{HAB}}}}{{{S_{ABC}}}} = \dfrac{{HC'}}{{CC'}}\)

Do đó: \(\dfrac{{{S_{HBC}} + {S_{HAC}} + {S_{HAB}}}}{{{S_{ABC}}}} \)\(\,= \dfrac{{HA'}}{{AA'}} + \dfrac{{HB'}}{{BB'}} + \dfrac{{HC'}}{{CC'}}\)

Hay \(1 = \dfrac{{HA'}}{{AA'}} + \dfrac{{HB'}}{{BB'}} + \dfrac{{HC'}}{{CC'}}\) (đpcm)


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