Question

cho ΔABC, Â=90*, AH ⊥ BC (H∈BC), cm:
a) AB.AC = BC.AH

b) AH² = BH.CH

Answer 1

Tự vẽ hình nha !

Câu a : Ta có :

\(\left\{{}\begin{matrix}S_{ABC}=\dfrac{1}{2}.AB.AC\\S_{ABC}=\dfrac{1}{2}.AH.BC\end{matrix}\right.\Rightarrow\dfrac{1}{2}.AB.AC=\dfrac{1}{2}.AH.BC\Rightarrow AB.AC=AH.BC\)

Câu b : Ta sẽ dễ dàng chứng minh được :

\(\left\{{}\begin{matrix}\Delta HBA\sim\Delta ABC\left(g-g\right)\\\Delta HAC\sim\Delta ABC\left(g-g\right)\end{matrix}\right.\Rightarrow\Delta HBA\sim\Delta HAC\left(g-g\right)\)

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