Prove:

( 1)∵∠BAC=∠DAE

∴∠BAC+∠CAD=∠DAE+∠CAD

That is ∠BAD=∠CAF.

In △ABD and △ACE:

AB=AC

Bad = coffee

AD=AE

∴△ABD≌△ACE(SAS)

∴BD=CE

(2) the intersection point a is AH⊥BD at point h and AG⊥CE at point g.

∵? △ABD?△ACE

∴S△ABD=S△ACE

∴ 1/2BD？ AH= 1/2CE？ Adjutant general

BD = CE

∴AH=AG

In APH and APG.

AH=AG

AP=AP

∴△APH≌△APG(HL)

∴∠APH=∠APG

AP splitting ∠BPE

(3) Take point O on PE, make OE=PD, and then connect with AO.

∫△ABD?△ACE

∴∠AEC=∠ADP

At LAPD and AOE.

AD=AE

∠AEC=∠ADP

PD=OE

∴△APD≌△AOE(SAS)

∴AO=AP

In △ANE and △PND.

∠∠AEC =∠ADP and ∠ANE=∠PND.

∴∠EPD=∠EAN=α

That is, ∠ EPD = 60.

∴∠APB+∠APE= 120

Once again ≈APB =∠APE

∴△APO is an equilateral triangle.

∴AP=PO

∴AP+PD=PO+OE

That is AP+PD=PE.