\(\Delta ABC\)vuông tại A có AB = 12cm; BC = 20cm, đường cao AH và trung tuyến AM, đườ...">
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29 tháng 3 2018

a)  Xét   \(\Delta HAC\) và     \(\Delta MAH\)có:

\(\widehat{AHC}=\widehat{AMH}=90^0\)

\(\widehat{HAC}\)      CHUNG

suy ra:   \(\Delta HAC~\Delta MAH\)

\(\Rightarrow\)\(\frac{AH}{AM}=\frac{AC}{AH}\)\(\Rightarrow\)\(AH^2=AM.AC\)

6 tháng 5 2020

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6 tháng 5 2020

ABCHKIEF

a) 

Xét \(\Delta\)ABC và \(\Delta\)HBA có: 

^BAC = ^BHA ( = 90 độ ) 

^ABC = ^HBA ( ^B chung ) 

=> \(\Delta\)ABC ~ \(\Delta\)HBA 

b) AB = 3cm ; AC = 4cm 

Theo định lí pitago ta tính được BC = 5 cm 

Từ (a) => \(\frac{AB}{BH}=\frac{BC}{AB}\Rightarrow BH=\frac{AB^2}{BC}=1,8\)

c) Xét \(\Delta\)AHC và \(\Delta\)AKH có: ^AKH = ^AHC = 90 độ 

và ^HAC = ^HAK ( ^A chung ) 

=> \(\Delta\)AHC ~ \(\Delta\)AKH 

=> \(\frac{AH}{AK}=\frac{AC}{AH}\Rightarrow AH^2=AC.AK\)

d) Bạn kiểm tra lại đề nhé!