Question

Cho tam giác ABC vuông ở A.Kẻ ba đường cao AH,BK,CM

a) CM: AB.AC=AH.BC?

b) CM: AC²=HC.BC

c) AH²=HB.HC

Answer 1

a) Xét ΔABH và ΔABC ta có:

\(\widehat{AHB}=\widehat{BAC}\)

\(\widehat{B}\) chung

→ΔABH ∼ ΔABC(g-g)(1)

\(\rightarrow\dfrac{AB}{AH}=\dfrac{BC}{AC}\)

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

b) Vì ΔABH ∼ ΔABC (cmt)

\(\rightarrow\dfrac{AC}{HC}=\dfrac{BC}{AC}\)

\(\rightarrow AC.AC=HC.BC\)

\(\Rightarrow AC^2=HC.BC\)

c) Xét ΔAHC và ΔABC ta có:

\(\widehat{C}\) chung

\(\widehat{AHC}=\widehat{BAC}=90^0\)

→ΔAHC ∼ ΔABC(g-g)(2)

Từ (1) và (2)→ΔABH ∼ ΔAHC

\(\rightarrow\dfrac{AH}{HB}=\dfrac{HC}{AH}\)

\(\rightarrow AH.AH=HB.HC\)

\(\Rightarrow AH^2=HB.HC\)