$(1) $证明：由题意得$\triangle AC'D≌\triangle ACD$
∴$∠ADC' = ∠ADC，$$∠AC'D = ∠C，$$C'D = CD$
又$∠ADC = 60°，$$∠BDC'+∠ADC'+∠ADC = 180°$
∴$∠BDC' = 60°$
在边$C'D$上取一点$P，$使$P D = BD，$连接$P B$
则$\triangle BDP $是等边三角形
∴$BP = BD = P D，$$∠P BD = ∠BP D = 60°$
又$BD=\frac 12CD，$∴$BD = \frac 12C'D，$
即$P C' = P D = BP$
∴$∠BC'P=∠P BC'$
又$∠BP D=∠BC'P+∠P BC'$
∴$∠BC'P=∠P BC'=\frac 12∠BP D = 30°$
即$∠C'BD=∠P BC'+∠P BD = 90°$
∴$BC'\perp BC$
$(2) $解：过点$A$分别作$BC，$$C'D，$$BC'$的垂线，
垂足分别为$E，$$F，$$G$
由$(1)$得$∠AC'D = ∠C，$$∠BC'P = 30°，$
$∠ADC' = ∠ADC，$$∠C'BD = 90°$
∴$DA$平分$∠CDC，$即$AE = AF$
又$∠C'BD=∠ABC+∠ABC'，$$∠ABC = 45°$
∴$∠ABC' = 45°，$即$∠ABC'=∠ABC$
∴$AB$平分$∠C'BD，$即$AE = AG$
∴$AF = AG$
在$Rt\triangle AC'F $和$Rt\triangle AC'G $中
$\begin {cases}AG = AF\\AC' = AC'\end {cases}$
∴$Rt\triangle AC'F≌ Rt\triangle AC'G(\mathrm {HL})$
∴$∠AC'F=∠AC'G$
又$∠BC'P+∠AC'F+∠AC'G = 180°$
∴$∠AC'F=\frac 12(180°-∠BC'P)=75°，$
即$∠C = 75°$
