数学求证
以△ABC的AB,AC为边向外作正△ABD,正△ACE,连BE,CD交于点P,求证:PB+PC+2PA=PD+PE.
以△ABC的AB,AC为边向外作正△ABD,正△ACE,连BE,CD交于点P,求证:PB+PC+2PA=PD+PE. 证明: ∵AD=AB,AC=AE,∠CAD=60+∠BAC=∠EAB, ∴△CAD≌△EAB,∴CD=BE,∠ADC=∠ABE, ∴A,D,B,P四点共圆,∠APD=∠ABD=60° 过A作∠PAK=60°,AK交PD于K, 易证△APB≌△AKD,∴KD=PB, ∴CD=PC+PB+PK=PC+PB+AP, 2CD=2PC+2PB+2AP, 又2CD=CD+BE=PC+PB+PD+PE, ∴2PC+2PB+2AP,=PC+PB+PD+PE, ∴PB+PC+2PA=PD+PE
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