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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!

8 tháng 4 2022

8 tháng 4 2022

ủa lớp 5 lm lớp 8

a: Xét ΔABC vuông tại A và ΔHBA vuông tại H có

góc ABC chung

Do đó: ΔABC\(\sim\)ΔHBA

Suy ra: BA/BH=BC/BA

hay \(BA^2=BH\cdot BC\)

b: Xét ΔCAB vuông tại A và ΔCHA vuông tại H có

góc ACB chung

Do đó: ΔCAB\(\sim\)ΔCHA
Suy ra: CA/CH=CB/CA

hay\(CA^2=CH\cdot CB\)

 

4 tháng 4 2023

Cậu ơi, cậu hk lm câu c cho tớ hả :3?

a: Xét ΔABH vuông tại H và ΔCBA vuông tại A có

góc B chung

=>ΔABH đồng dạng với ΔCBA

b: ΔABC vuông tại A

mà AH là đường cao

nên HA^2=HB*HC

c: AI/IH=BA/BH

EC/AE=BC/BA

mà BA/BH=BC/BA

nên AI/IH=EC/AE
=>AI*AE=IH*EC

a: Xét ΔAMB vuông tại M và ΔANC vuông tạiN có

góc A chung

=>ΔAMB đồng dạng vơi ΔANC

=>AM/AN=AB/AC

=>AM*AC=AB*AN; AM/AB=AN/AC

b: Xét ΔAMN và ΔABC có

AM/AB=AN/AC
góc A chung

=>ΔAMN đồng dạng với ΔABC

=>góc AMN=góc ABC

a: \(\text{Δ}ABC\sim\text{Δ}HBA;\text{Δ}ABC\sim\text{Δ}HCA\)

b: \(BC=\sqrt{AB^2+AC^2}=25\left(cm\right)\)

\(AH=\dfrac{AB\cdot AC}{BC}=\dfrac{15\cdot20}{25}=12\left(cm\right)\)

\(BH=\dfrac{AB^2}{BC}=\dfrac{15^2}{25}=9\left(cm\right)\)

CH=BC-BH=25-9=16(cm)