Question

Theo hình vẽ hãy:

a) AB² = BH . BC

b) AC² = CH.BC

c) AH² = HB.HC

d) AB.AC=BC.AH

e) BC² = AB² + AC²

f) \(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)

Answer 1

B. Xin lỗi nãy mình nhầm

Xét \(\Delta ABCvà\Delta HAC\) có:

góc A = góc H ( =90^o)

góc C chung

Do đó: \(\Delta ABC\sim\Delta HAC\) (g.g)

\(\Rightarrow\dfrac{AC}{BC}=\dfrac{CH}{AC}\Rightarrow AC^2=BC.CH\) ( ĐPCM)

Answer 2

a.

Xét \(\Delta ABH\) và \(\Delta CBA\) có:

góc H bằng góc A(=90^o)

góc B chung

Do đó: \(\Delta ABH\sim CBA\) (g-g)

\(\Rightarrow\dfrac{AB}{BH}=\dfrac{BC}{AB}\Rightarrow AB^2=BC.BH\) ( ĐPCM)

Answer 3

b.

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

góc A = góc H ( =90^o)

góc C chung

Do đó: \(\Delta ABC\sim\Delta HBC\)

\(\Rightarrow\dfrac{AC}{BC}=\dfrac{CH}{AC}\) \(\Rightarrow AC^2=BC.CH\) (ĐPCM)

Answer 4

Ta có: \(\Delta ABH\sim\Delta CBA\)

=> góc BAH = góc ACB

Xét \(\Delta ABH\) và \(\Delta CAH\)

có: góc H = 90^o

góc BAH = góc ACB (cmt)

Do đó: \(\Delta ABH\sim\Delta CAH\)

\(\Rightarrow\dfrac{AH}{HB}=\dfrac{HC}{AH}\Rightarrow AH^2=HB.HC\) ( D0PCM)

Answer 5

d.

Ta có: \(\Delta ABH\sim\Delta CBA\)

\(\Rightarrow\dfrac{AB}{AH}=\dfrac{BC}{AC}\Rightarrow AB.AC=BC.AH\) ( ĐPCM)

e.

Ta có tam giác ABC vuông tại A

=> BC² = AB² + AC² ( định lý Pytago)

f. Ta có: \(\Delta ABH\sim\Delta CBA\)

\(\Rightarrow\dfrac{AH}{AB}=\dfrac{AC}{BC}\Rightarrow AH.BC=AB.AC\)

\(\Rightarrow AH^2.BC^2=AB^2.AC^2\)

\(\Leftrightarrow AH^2\left(AB^2+AC^2\right)=AB^2.AC^2\)

\(\Leftrightarrow\dfrac{1}{AH^2}=\dfrac{AB^2+AC^2}{AB^2.AC^2}\)

\(\Leftrightarrow\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)