a, Xét[tex]\Delta BDC và \Delta HBC[/tex] có
[tex]\angle C[/tex] chung
[tex]\angle BDC = \angle BHC (=90^{o})[/tex]
=>[tex]\Delta BDC = \Delta HBC[/tex]
b, Vì [tex]\Delta BDC = \Delta HBC[/tex]
=>[tex]\frac{BC}{CD}=\frac{HC}{BC}[/tex]
=> [tex]HC=\frac{BC^{2}}{CD}[/tex]
=>[tex]HC=\frac{15^{2}}{25}[/tex]
=> HC=9(cm)
=> HD=CD-HC=25-9=16(cm)