Question

[Dangthybgggg]

Cho tam giác ABC vuông tại A, AH là đường cao

Cm: a, AB²=BC.BH

b, AC²=BC.HC

c, AB. AC=AH. BC

d, AH²=BH.HC

Answer 1

[Boul]

a) \(\text{Xét }\Delta\text{ vuông }HBA\text{ và }\Delta\text{ vuông }ABC,\text{có: }\)

\(\widehat{BHA}=\widehat{BAC}=90^o\)

\(\widehat{ABC}\text{ chung }\)

\(\Rightarrow\Delta\text{ vuông }HBA\text{ đồng dạng }\Delta\text{ vuông }ABC\)

\(\Rightarrow\frac{HB}{AB}=\frac{HA}{AC}=\frac{AB}{BC}\)

\(\frac{BH}{AB}=\frac{AB}{BC}\Leftrightarrow BC.HB=AB^2\)

b) cm tương tự câu a:

t/g vuông BAC đồng dạng t/g vuông AHC

\(\Rightarrow\frac{AB}{AH}=\frac{BC}{AC}=\frac{AC}{HC}\)

\(\frac{BC}{AC}=\frac{AC}{HC}\Leftrightarrow AC^2=BC.HC\)

c) \(\frac{HA}{AC}=\frac{AB}{BC}\left(\text{câu a}\right)\)

\(\Leftrightarrow HA.BC=AC.AB\)

d) \(\left\{{}\begin{matrix}\frac{AH}{AC}=\frac{HB}{AB}\\\frac{AB}{AH}=\frac{AC}{HC}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AH=\frac{HB.AC}{AB}\\AH=\frac{AB.HC}{AC}\end{matrix}\right.\)

\(\Leftrightarrow AH^2=\frac{HB.AC}{AB}\cdot\frac{AB.HC}{AC}=HB.HC\)