Question

cho tam giac nhon ABC, cac duong cao AD, BE, CF cắt nhau tại H.chứng minh:

a.tam giác AEF đồng dạng tam giác ABC

b.AD.HD=BD.DC

c.AH.HD=BH.HE=CH.HF

d.BE.BH+CE.CH=BC²

Answer 1

B đã có câu tl chưa ạ

Answer 2

a) Xet tam giac AEB va tam giac AFC  co:AE/AF =AB/AC suy za AE/AB=AF/AC(1)

(2) chung goc A

Tu (1) va (2) suy za tam giac AFE~ACB(CGC)

May y cn laj mjnk ngaj nghj. Mong ban thong cam

Answer 3

d) Xet tg BHD ~BCE(gg)=> BH/BC=BD/BE=>BH.BE=BD.BC(1)

Xet tg CHD~CBF(gg) =>CH/CB=CD/CF=>CH.CF=CB.CD(2)

Tu (1) va (2) => BH.BE+CH.CF=BD.BC+BC.CD=>BC(BD=CD)=>BC.BC=BC^2

Answer 4

a) Ta có \(\Delta AEB\sim\Delta AFC\left(g-g\right)\Rightarrow\frac{AE}{AF}=\frac{AB}{AC}\Rightarrow\frac{AE}{AB}=\frac{AF}{AC}\)

\(\Rightarrow\Delta AEF\sim\Delta ABC\left(g-g\right)\)

b) Ta có \(\Delta ABD\sim\Delta CBF\left(g-g\right)\Rightarrow\widehat{BAD}=\widehat{BCF}\)

\(\Rightarrow\Delta ABD\sim\Delta CHD\left(g-g\right)\Rightarrow\frac{AD}{CD}=\frac{BD}{HD}\)

\(\Rightarrow AD.HD=CD.BD\)

c) Ta có \(\Delta AEH\sim\Delta BDH\left(g-g\right)\Rightarrow\frac{AH}{BH}=\frac{HE}{HD}\Rightarrow AH.HD=HB.HE\)

\(\Delta HDC\sim\Delta HFA\left(g-g\right)\Rightarrow\frac{HD}{HF}=\frac{HC}{HA}\Rightarrow AH.HD=HF.HC\)

\(\Rightarrow AH.HD=BH.HE=CH.HF\)

d)  Ta có ngay \(\Delta BDH\sim\Delta BEC\left(g-g\right)\Rightarrow\frac{BD}{BE}=\frac{BH}{BC}\Rightarrow BE.BH=BC.BD\)

Tương tự \(CE.CH=BC.DC\)

\(\Rightarrow BE.BH+CE.CH=BC.BD+BC.DC=BC^2\)