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证明：∵​$AB = AC，$​​$BD = CE$​
∴​$AB + BD = AC + CE，$​即​$AD = AE$​
​$ $​在​$\triangle ACD$​和​$\triangle ABE$​中
​$\begin {cases}AC = AB\\∠CAD = ∠BAE\\AD = AE\end {cases}$​
∴​$\triangle ACD≌\triangle ABE(S AS)$​
∴​$∠D = ∠E$​
​$ $​在​$\triangle BDF $​和​$\triangle CEF $​中
​$\begin {cases}∠D = ∠E\\∠BF D = ∠CFE\\BD = CE\end {cases}$​
∴​$\triangle BDF≌\triangle CEF(\mathrm {AAS})$​
∴​$BF = CF$​
​$ $​在​$\triangle ABF $​和​$\triangle ACF $​中
​$\begin {cases}AB = AC\\BF = CF\\AF = AF\end {cases}$​
∴​$\triangle ABF≌\triangle ACF(\mathrm {SSS})$​
∴​$∠BAF = ∠CAF，$​即​$AF $​平分​$∠DAE$​
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证明：∵​$∠ABC = 90°，$​​$∠ABC + ∠DBE = 180°$​
∴​$∠DBE = 180°-∠ABC = 90°$​
在​$Rt\triangle ACB$​和​$Rt\triangle DEB$​中
​$\begin {cases}AC = DE\\BC = BE\end {cases}$​
∴​$Rt\triangle ACB≌ Rt\triangle DEB(\mathrm {HL})$​
∴​$AB = DB$​
过点​$B$​作​$BM$​平分​$∠ABD，$​交​$AK$​于点​$M，$​则​$∠ABM = ∠MBD = 45°$​
又​$KB$​平分​$∠AKG，$​∴​$∠AKB = ∠BKG$​
∵​$BF $​平分​$∠ABC，$​∴​$∠ABF = ∠CBF = 45°$​
∵​$∠CBF = ∠DBG，$​∴​$∠DBG = 45°$​
∴​$∠ABM = ∠MBD = ∠DBG$​
在​$\triangle BMK$​和​$\triangle BGK$​中
​$\begin {cases}∠MBK = ∠G BK\\BK = BK\\∠MKB = ∠GKB\end {cases}$​
∴​$\triangle BMK≌\triangle BGK(AS A)$​
∴​$BM = BG，$​​$KM = KG$​
在​$\triangle ABM$​和​$\triangle DBG $​中
​$\begin {cases}AB = DB\\∠ABM = ∠DBG\\BM = BG\end {cases}$​
∴​$\triangle ABM≌\triangle DBG(S AS)$​
∴​$AM = DG$​
∵​$AK = AM + KM，$​∴​$AK = DG + KG$​
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