Question

Cho tam giác ABC vuông tại A,đường cao AH(H thuộc BC) và phân giác BE của góc ABC(E thuộc AC) cắt nhau tại I.Chứng minh:

a) IH.AB=IA.BH

b) Tam giác BHA ~ tam giác BAC suy ra AB^2 = BH.BC

c) IH/IA = AE/EC

d) Tam giác AIE cân

Answer 1

xet \(\Delta BHI\) va \(\Delta BAE\) co

\(\widehat{BAE}=\widehat{BHI}=90^0\)va \(\widehat{ABE}=\widehat{IBH}\) (BE la pg)

\(\Rightarrow\Delta BHI\simeq\Delta BAE\left(gg\right)\)

\(\Rightarrow\dfrac{BH}{AB}=\dfrac{IH}{AE}\)

D,Ta co: \(\widehat{AIE}=\widehat{BIH}\left(dd\right)\)

ma \(\widehat{BIH}=\widehat{BEA}\left(\Delta BHI\simeq\Delta BAE\right)\)

\(\Rightarrow\widehat{AIE}=\widehat{BEA}\Rightarrow\Delta AIE\) can tai A

\(\Rightarrow AI=AE\)

A,\(\Rightarrow\dfrac{BH}{AB}=\dfrac{IH}{IA}\Rightarrow BH.IA=AB.IH\)

B, xet \(\Delta BHA\) va \(\Delta BAC\) co

\(\widehat{B}\) chung, \(\widehat{BAE}=\widehat{BHA}=90^0\)

\(\Rightarrow\Delta BHA\simeq\Delta BAC\left(gg\right)\)

C, Vi \(\Delta BHI\simeq\Delta BAE\)

\(\Rightarrow\dfrac{IH}{AE}=\dfrac{BH}{AB}\left(1\right)\)

Vi \(\Delta BHA\simeq\Delta BAC\)

\(\Rightarrow\dfrac{AH}{AC}=\dfrac{BH}{BA}\left(2\right)\)

Tu (1) va (2)\(\Rightarrow\dfrac{HI}{AE}=\dfrac{AH}{AC}\Rightarrow\dfrac{HI}{AH}=\dfrac{AE}{AC}\)

\(\Rightarrow\dfrac{HI}{AH-HI}=\dfrac{AE}{AC-AE}\Rightarrow\dfrac{IH}{IA}=\dfrac{AE}{EC}\)

cai nay \(\simeq\) la dong dang do nha bn

Answer 2

Answer 3