$解：?(1)?在?Rt△ABC?中，∵?BC=6，??AB=10?$
$∴?AC=\sqrt {AB^2-BC^2}=8?$
$∴?tan A=\frac {BC}{AC}=\frac 34?$
$∵?CD?是斜邊上的高$
$∴?∠CDB=∠ACB=90°?$
$∴?∠B+∠BCD=∠ACD+∠BCD=90°?$
$∴?∠B=∠ACD?$
$∴?tan ∠ACD=tan B=\frac {AC}{BC}=\frac 43?$
$?(2)?不妨設?AD=9x，?則?BD=4x?$
$∵?∠ACD=∠B?$
$∴?90°-∠ACD=90°-∠B，?即?∠BCD=∠A?$
$∴?tan ∠BCD=tan A，?即?\frac {BD}{CD}=\frac {CD}{AD}?$
$∴?CD^2=BD · AD?$
$∵?AD=9x，??BD=4x?$
$∴?CD=6x?$
$∴?tan ∠BCD=\frac {BD}{CD}=\frac {4x}{6x}=\frac 23?$
