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.
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