$解：?(2)?過點?D?作?DE//AC?交?BC?的延長線于點?E?，如圖$
$∵?AD//BC(?已知?)?，即?AD//CE?$
$∴四邊形?ACED?是平行四邊形$
$∴?AC = DE?$
$又∵四邊形?ABCD?是等腰梯形$
$∴?AC = BD?$
$∴?BD = DE?$
$∵?DE//AC?，?AC⊥BD?$
$∴?∠BDE = ∠BPC = 90°?$
$∴?∠DBC = 45°?$
$∴?PC = PB = 7\ \mathrm {cm}?$
$∵?AD//BC?$
$∴?∠ADP = 45°?$
$∴?PD = PA = 3\ \mathrm {cm}?$
$∴?AC = BD = 7+3 = 10\ \mathrm {cm}?$
$∴?S_{梯形ABCD}= \frac 12AC·BD=\frac 12×10×10= 50(\mathrm {cm}^2)?$
