解：$(1)$∵$AC = AF，$∴$∠ACF=∠AF C$
$ $又$∠CAF=α，$$∠ACF+∠AF C+∠CAF = 180°$
∴$∠AF C=∠ACF=\frac 12(180°-∠CAF)=90°-\frac 12α$
$ (2)$由题意，得$∠BAC=∠ACB = 60°，$$AB = BC = AC = AF$
∴$∠AF B=∠ABF=\frac 12(180°-∠BAF)=\frac 12(180°-∠BAC-∠CAF)$
$=90°-\frac 12∠BAC-\frac 12∠CAF=60°-\frac 12∠CAF$
$ $由$(1)$得$∠AF C=∠ACF=\frac 12(180°-∠CAF)=90°-\frac 12∠CAF$
∴$∠GFH=∠AF C-∠AF B=90°-\frac 12∠CAF-(60°-\frac 12∠CAF)=30°$
$ (3)$如图，连接$BG，$$CH，$过点$C$作$CM\perp BF $于点$M$
$ $由$(2)$得$∠ACB = 60°，$$∠GFH = 30°，$$AB = BC = AC = AF，$且$AH\perp BF$
∴$GH=\frac 12FG，$$∠FGH = 90°-∠GFH = 60°$
∵$C$为$GF $的中点，∴$CH = CF = CG=\frac 12GF，$即$GH = CG = CH$
∴$\triangle CGH$为等边三角形。∴$∠G CH=∠ACB = 60°$
∴$∠G CH-∠BCH=∠ACB-∠BCH，$即$∠G CB=∠HCA$
∴$\triangle G CB≌\triangle HCA(S AS)，$∴$AH = BG$
$ $设$GH = CG = CH = CF = a，$则$GF = 2a$
∵$AB = AF，$$AH\perp BF，$∴$AG $垂直平分$BF$
∴$BG = GF = 2a，$即$AH = BG = 2a$
同理，得$CM=\frac 12CF=\frac 12a$
∵$S_{\triangle ADH}=\frac 12\ \mathrm {A}H·DH，$$S_{\triangle CDH}=\frac 12CM·DH，$
∴$\frac {S_{\triangle ADH}}{S_{\triangle CDH}}=\frac {\frac 12\ \mathrm {A}H·DH}{\frac 12CM·DH}=\frac {AH}{CM}=4$
同理，得$\frac {S_{\triangle ADH}}{S_{\triangle CDH}}=\frac {AD}{CD}，$则$\frac {AD}{CD}=4$
∵$CD = 1.5，$∴$AD = 6$
