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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!

4 tháng 4 2017

Theo câu a) ta có: \(AH^2=AI.AB\left(1\right)\)

Xét tam giác AHK và tam giác ACH có:

góc A chung; góc AKH = góc AHC = 900

=> tam giác AHK đồng dạng với tam giác ACH (g-g)

=>\(\dfrac{AK}{AH}=\dfrac{AH}{AC}\Rightarrow AK.AC=AH^2\left(2\right)\)

Từ (1)(2) => \(AI.AB=AK.AC\Rightarrow\dfrac{AI}{AC}=\dfrac{AK}{AB}\)

Xét tam giác AIK và tam giác ABC có:

góc A chung; \(\dfrac{AI}{AC}=\dfrac{AK}{AB}\)

=> Tam giác AIK đồng dạng với tam giác ACB (c-g-c)

3 tháng 4 2017

a) Xét tam giác AIH và tam giác AHB có:

góc BAH chung; góc AIH = góc AHB (= 900)

=> tam giác AIH = tam giác AHB (g-g)

\(\Rightarrow\dfrac{AH}{AI}=\dfrac{AB}{AH}\Rightarrow AH^2=AI.AB\)

1 tháng 4 2019

a) Xét tam giác ABC và tam giác HBA có Góc ABC chungg,góc BHA=góc BAC=90 độ

=> Tam giác ABC đồng dạng với tam giác HBA(gg)=> \(\frac{AB}{HB}=\frac{BC}{AB}\)=> AB^2=BH.BC

1 tháng 4 2019

b)Tam giác ABC có BF là phân giác góc ABC=>\(\frac{BC}{AB}=\frac{FC}{AF}\)mà \(\frac{AB}{HB}=\frac{BC}{AB}\)=>\(\frac{AB}{BH}=\frac{FC}{AF}\left(1\right)\)

Tam giác ABH có BE là phân giác goc ABH =>\(\frac{BA}{BH}=\frac{AE}{EH}\left(2\right)\)

Từ 1 và 2=>\(\frac{FC}{AF}=\frac{AE}{EH}=>\frac{EH}{AE}=\frac{AF}{FC}\)

26 tháng 4 2021

a) Xét \(\Delta CEF\)và \(\Delta CAB\)có:

\(\widehat{CFE}=\widehat{CBA}\left(=90^0\right)\).

\(\widehat{BCA}\)chung.

\(\Rightarrow\Delta CEF~\Delta CAB\left(g.g\right)\)(điều phải chứng minh).

26 tháng 4 2021

b) Xét \(\Delta ABC\)và \(\Delta FBK\)có:

\(\widehat{KBC}\)chung.

\(\widehat{BAC}=\widehat{BFK}\left(=90^0\right)\).

\(\Rightarrow\Delta ABC~\Delta FBK\left(g.g\right)\).

\(\Rightarrow\frac{BA}{BF}=\frac{BC}{BK}\)(tỉ số đồng dạng).

\(\Rightarrow BA.BK=BF.BC\)(điều phải chứng minh).