$(1)$证明：过点$P$作$PD\bot BC$于$D$，
$\because \angle ABC$和$\angle ACB$的角平分线相交于点$P$，且$PE\bot AB$，$PF\bot AC$，
$\therefore PD=PE$，$PD=PF$，
$\therefore PE=PF$；
$(2)$$\because PE=PF$，$PE\bot AB$，$PF\bot AC$，
$\therefore AP$平分$\angle BAC$，
同理可得，$BP$是$\angle ABC$的平分线，$CP$是$\angle ACB$的平分线，
$\because \angle BAC=60^{\circ}$，
$\therefore \angle ABC+\angle ACB=120^{\circ}$，
$\therefore \angle PBD+\angle PCD=\frac{1}{2}×120°=60^{\circ}$，
$\therefore \angle BPC=180^{\circ}-60^{\circ}=120^{\circ}$.

$(1)$证明：过点$P$作$PD\bot BC$于$D$，
$\because \angle ABC$和$\angle ACB$的角平分线相交于点$P$，且$PE\bot AB$，$PF\bot AC$，
$\therefore PD=PE$，$PD=PF$，
$\therefore PE=PF$；
$(2)$$\because PE=PF$，$PE\bot AB$，$PF\bot AC$，
$\therefore AP$平分$\angle BAC$，
同理可得，$BP$是$\angle ABC$的平分线，$CP$是$\angle ACB$的平分线，
$\because \angle BAC=60^{\circ}$，
$\therefore \angle ABC+\angle ACB=120^{\circ}$，
$\therefore \angle PBD+\angle PCD=\frac{1}{2}×120°=60^{\circ}$，
$\therefore \angle BPC=180^{\circ}-60^{\circ}=120^{\circ}$.