a) Xét \(\Delta HBA\)và \(\Delta ABC\)có :

\(\widehat{AHB}=\widehat{BAC}=90^o;\widehat{B}\left(chung\right)\)

\(\Rightarrow\)\(\Delta HBA\)\(\approx\)\(\Delta ABC\)( g.g )

b) Xét \(\Delta HBA\)và \(\Delta HAC\)có :

\(\widehat{AHB}=\widehat{AHC}=90^o\)

\(\widehat{BAH}=\widehat{ACH}\left(cung-phu-\widehat{B}\right)\)

\(\Rightarrow\Delta HBA\approx\Delta HAC\left(g.g\right)\)

\(\Rightarrow\frac{BH}{AH}=\frac{AH}{HC}\Rightarrow AH^2=BH.HC\)

a)Xét \(\Delta ABC\)và \(\Delta HBA\)có:

\(\widehat{BAC}=\widehat{BHA}\)(=\(90^0\))

\(\widehat{B}\)chung

=>\(\Delta ABC\)~\(\Delta HBA\)(g.g)

=>\(\dfrac{AB}{HB}=\dfrac{BC}{AB}\)

=>\(AB^2=HB.BC\)

a) Ta có:   \(\widehat{HAB}+\widehat{HBA}=90^0\)

                 \(\widehat{HAB}+\widehat{HAC}=90^0\)

suy ra:   \(\widehat{HBA}=\widehat{HAC}\)

Xét 2 tam giác vuông:  \(\Delta HBA\) và  \(\Delta HAC\) có:

           \(\widehat{BHA}=\widehat{AHC}=90^0\)

          \(\widehat{HBA}=\widehat{HAC}\)   (CMT)

suy ra:   \(\Delta HBA~\Delta HAC\)

b)   \(BC=BH+HC=25+36=61\)cm

 \(\Delta HBA~\Delta HAC\) \(\Rightarrow\)\(\frac{HB}{HA}=\frac{AB}{AC}\)

\(\Rightarrow\)\(\frac{AB}{AC}=\frac{5}{6}\)\(\Leftrightarrow\)\(\frac{AB}{5}=\frac{AC}{6}\)\(\Leftrightarrow\)\(\frac{AB^2}{25}=\frac{AC^2}{36}=\frac{AB^2+AC^2}{25+36}=\frac{BC^2}{61}=\frac{61^2}{61}=61\)

suy ra:    \(\frac{AB^2}{25}=61\) \(\Leftrightarrow\) \(AB=\sqrt{1525}\) cm

            \(\frac{AC^2}{36}=61\)\(\Leftrightarrow\) \(AC=\sqrt{2196}\)cm

p/s: tham khảo

a: Xét ΔHBA vuông tại H và ΔHAC vuông tại H có

\(\widehat{HBA}=\widehat{HAC}\)

Do đó: ΔHBA\(\sim\)ΔHAC

b: \(BC=HB+HC=61\left(cm\right)\)

\(AB=\sqrt{25\cdot61}=5\sqrt{61}\left(cm\right)\)

\(AC=\sqrt{36\cdot61}=6\sqrt{61}\left(cm\right)\)

a.

 Xét tam giác ABH và tam giác CBA có : góc BHA = góc BAC (2 góc = 90 độ )

                                                            góc ABH = góc CBA (2 góc chung )

Suy ra tam giác ABH đồng dạng với tam giác CBA ( trường hợp g.g )

b.

Xét tam giác ABH có BI là phân giác góc ABH suy ra  AI\AH = BA\BH

Suy ra AI.BH = IH . BA