Question

[Vũ Trọng Hoàn]

Cho\(\Delta ABC\)nhọn các đường cao BD và CE cắt nhauowr H.Chứng minh rằng:

a)\(\Delta BAD~\Delta CAE\)

b)HB.HD=HC.HE

c)\(\Delta BHC~\Delta DHE\)

d)DH.DB=DA.DC

Answer 1

[❥一Ǥoɗɗεşş øf iηէεℓℓεcէ︵♛╾━╤デ╦︻]

a, xét \(\Delta BAD\)và\(\Delta CAE\)có:

         \(\widehat{BAD}=\widehat{CAE}\)(góc chung)

        \(\widehat{BDA}=\widehat{CEA}\left(90^0\right)\)

\(\Rightarrow\Delta BAD~\Delta CAE\left(g.g\right)\)

\(b,\)xét \(\Delta EHB\)và\(\Delta DHC\)có:

          \(\widehat{EHB}=\widehat{DHC}\)(đối đỉnh)

          \(\widehat{HEB}=\widehat{HDC}\left(=90^0\right)\)

\(\Rightarrow\Delta EHB~\Delta DHC\left(g.g\right)\)

\(\Rightarrow\frac{HE}{HD}=\frac{HB}{HC}\Rightarrow HE.HC=HB.HD\)

c,\(HE.HC=HB.HD\Rightarrow\frac{HE}{HB}=\frac{HD}{HC}\)

xét\(\Delta EHD\)và\(\Delta BHC\)có

        \(\frac{HE}{HB}=\frac{HD}{HC}\left(cmt\right)\)

         \(\widehat{EHD}=\widehat{BHC}\)(đối đỉnh)

\(\Rightarrow\Delta EHD~\Delta BHC\left(c.g.c\right)\)