$解：?(1)?過(guò)?P ?作?P E⊥A B?，?P F⊥A C?$
$∵?AP ?平分?∠BAC?并交? BC?于點(diǎn)?P?$
$∴?P E ⊥ A B?，?P F⊥ A C?$
$∴?PE=PF?$
$∴?S_{\triangle ABP}?與?S_{\triangle ACP}?的比? =\frac {1}{2}\ \mathrm {A} B ·P E∶ \frac {1}{2}\ \mathrm {A} C ·P F=A B∶ A C=5∶ 4 ?$
$?(2)?∵?\frac {S_{\triangle A B P}}{S_{\triangle A C P}}=\frac {\frac {1}{2}\ \mathrm {A} B ×P E}{\frac {1}{2}\ \mathrm {A} C ×P F}=\frac {P B}{P C}?$
$∴?\frac {P B}{P C}=\frac {A B}{A C}=\frac {5}{4}?$
$∴?P B=\frac {5}{9}\ \mathrm {B} C=\frac {10}{3}?$
