(1)证明：因为$AH\perp BC，$$\angle BAC = 90^{\circ}，$所以$\angle AHC = 90^{\circ}=\angle BAC，$所以$\angle BAH+\angle CAH = 90^{\circ}，$$\angle BAH+\angle B = 90^{\circ}，$所以$\angle CAH=\angle B。$
在$\triangle ABH$和$\triangle CAH$中，
$\begin{cases}\angle B=\angle CAH\\\angle AHB=\angle AHC\\AB = CA\end{cases}$
所以$\triangle ABH\cong\triangle CAH(AAS)，$所以$BH = AH，$$AH = CH，$所以$AH=\frac{1}{2}BC。$
$ (2) $解:
①如图，过点$A$作$AH\perp BP$于点$H，$连接$AP，$在$BP$上取一点$D$使得$BD = PC，$则$DP = BP - BD = 6 - 1 = 5。$
设$AC$与$BP$交于点$E。$
因为$\angle BAC=\angle BPC = 90^{\circ}$且$\angle AEB=\angle PEC，$所以$\angle ABD=\angle ACP。$
又因为$AB = AC，$$BD = CP，$所以$\triangle ADB\cong\triangle APC(SAS)，$所以$AD = AP，$$\angle BAD=\angle PAC，$所以$\angle DAP=\angle DAE+\angle PAC=\angle BAD+\angle DAE=\angle BAC = 90^{\circ}。$
因为$AH\perp DP，$所以$AH=\frac{1}{2}DP=\frac{5}{2}。$
②如图，过点$A$作$AH\perp BP$于点$H，$连接$AP，$在$PB$的延长线上取一点$D$使得$BD = PC，$则$DP = BP + BD = 6 + 1 = 7。$
因为$\angle BAC=\angle BPC = 90^{\circ}，$所以$\angle ABP+\angle ACP = 180^{\circ}，$又因为$\angle ABP+\angle ABD = 180^{\circ}，$所以$\angle ABD=\angle ACP。$
又因为$AB = AC，$$BD = CP，$所以$\triangle ADB\cong\triangle APC(SAS)，$所以$AD = AP，$$\angle BAD=\angle CAP，$所以$\angle DAP=\angle DAB+\angle BAP=\angle CAP+\angle BAP=\angle BAC = 90^{\circ}。$
因为$AH\perp DP，$所以$AH=\frac{1}{2}DP=\frac{7}{2}。$
