Question

Cho \(\Delta ABC\) nhọn,kẻ 2 đường cao BE và CF cắt nhau tại H.

a)\(\Delta ABE\sim\Delta ACF\)

b)CH.CF=BH.EH

c)Biết AE=6cm,AB=10cm.Tính S\(\Delta_{ÁBC}\)(Biết S\(\Delta_{AEF}=20cm^2\))

Answer 1

a) Xét \(\Delta ABE,\Delta ACF\) có :

\(\left\{{}\begin{matrix}\widehat{A}:Chung\\\widehat{AEB}=\widehat{AFC}=90^o\end{matrix}\right.\)

\(\Rightarrow\Delta ABE\sim\Delta ACF\left(g.g\right)\)

b) Xét \(\Delta BFH,\Delta CEH\) có :

\(\left\{{}\begin{matrix}\widehat{BFH}=\widehat{CEH}=90^o\\\widehat{BHF}=\widehat{CHE}\left(\text{Đối đỉnh}\right)\end{matrix}\right.\)

=> \(\Delta BFH\sim\Delta CEH\left(g.g\right)\)

\(\Rightarrow\dfrac{CH}{BH}=\dfrac{EH}{CF}\)

\(\Rightarrow CH.CF=BH.EH\)