$證明：(1)過點(diǎn)A作AH⊥BC，垂足為H$
$∵CD是△ABC的高$
$∴∠AHB=∠AHC=∠BDC=90°$
$∴∠BAH+ ∠ABC=90°，$
$∠BCD+∠ABC=90°$
$∴∠BAH= ∠BCD$
$∵∠BAC = 2 ∠BCD$
$∴∠BAC = 2∠BAH$
$∴∠BAH=∠CAH$
$在△ABH和△ACH 中$
$\begin{cases}∠AHB=∠AHC\\AH=AH\\∠BAH=∠CAH\end{cases}$
$∴△ABH ≌ △ACH$
$∴BH=CH∴$
$∵∠APC=60°$
$∴∠PAH=90°-60°=30°$
$∴PA=2PH$
$∵PB=PH-BH，PC=PH+HC$
$∴PB+PC=PH-BH+PH+CH$
$=2PH=PA$
