Question

Co tam giac ABC vuong tai B,duong cao BH,AB=3cmBC=4cm,ke phan giac BI.Tinh do dai AC,CI.CM:tam giac BACdong dang tam giac HBC.Ten tia doi tia BAlay diem D,ve BK vuong goc CD.CM:BC^2=CK xCD.Cho BD=7cm.Tinh dien tich tam giac CHK?

 

Answer 1

a) Ta có : Tam giác ABC vuông ở B

=> AB² + BC² = AC²

=> 3² + 4² = AC²

=> AC² = 25

=> AC = 5 (cm)

Vì BI là tia phân giác góc B

=> \(\frac{AI}{IC}=\frac{AB}{BC}\)

=> \(\frac{AI+IC}{IC}=\frac{AB+BC}{BC}\)

=> \(\frac{AC}{IC}=\frac{AB+BC}{BC}\)

=> \(IC=\frac{AC.BC}{AB+BC}=\frac{5.4}{3+4}=\frac{20}{7}\left(cm\right)\)

Answer 2

b) Xét tam giác ABC và tam giác HBC có

 \(\hept{\begin{cases}\widehat{ACB}\text{ chung }\\\widehat{CHB}=\widehat{CBA}=90^{\text{o}}\end{cases}}\)

=> \(\Delta BAC\approx\Delta HBC\left(g-g\right)\)(1)

c) Xét tam giác CBK  và tam giác CDB có : 

\(\hept{\begin{cases}\text{\widehat{D} Chung }\\\widehat{BKD}=\widehat{CBD}\left(=90^{\text{o}}\right)\end{cases}}\)\(\hept{\begin{cases}\widehat{C}\text{ chung }\\\widehat{CBD}=\widehat{BKC}\left(=90^{\text{o}}\right)\end{cases}}\)

=> \(\Delta CBK\approx\Delta CDB\left(g-g\right)\)

=> \(\frac{BC}{CD}=\frac{BK}{BD}=\frac{CK}{BC}\)

=> \(\frac{BC}{CD}=\frac{CK}{BC}\Rightarrow BC^2=CK.CD\)