初中几何一题，求角

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在$\triangle$ABC中,AB=AC,D在AC的延长线上,AD=BC,$BD=\sqrt2BC.$求$\angle A$的度数

三角法
设$\angle ABC=\angle ACB=x$,对$\triangle ABC$用余弦定理,$(2\sqrt2\cos x)^2=1+(2\cos x)^2+4\cos x\cos 2x,\cos x$是$8 x^3-4 x^2-4 x+1$的根.
$\cos\frac{3\pi}7=-\cos\frac{4\pi}7$,设$x=\cos\frac{\pi}7$,则$4x^3-3x+2(2x^2-1)^2-1=0,(x+1) \left(8 x^3-4 x^2-4 x+1\right)=0,$同理$\cos\frac{3\pi}7,\cos\frac{5\pi}7$满足$8 x^3-4 x^2-4 x+1=0$,它们是$8 x^3-4 x^2-4 x+1$的三个根.由BC>AC知$x<\frac\pi3$,因此$x=\frac{\pi}7$,$\angle A=\frac{900^\circ}7$.
几何法
取BC中点E,取BE上一点F使得BF=AE,设$\angle ABD=\angle ADB=x,\angle CBD=\alpha,$则$\angle C=\angle ADB-\angle CBD=x-\alpha,$又$\frac{BC}{AC}=\frac{BC}{BD}=\sqrt2$,E为BC中点$\Rightarrow\frac {AC}{CE}=\frac{BC}{AC}=\sqrt2,\frac{BD}{CE}=\frac{BC}{BD}=\sqrt2\Rightarrow\triangle ACE\sim\triangle BCA,\triangle BDE\sim\triangle BCD,\angle BDE=\angle C=x-\alpha,\angle EAC=\angle ABC=x+\alpha,\angle AEB=180^\circ-\angle AEC=180^\circ-\angle BAD=2x$,又BF=AE,BA=AE$\Rightarrow\triangle ABF\sim\triangle DAE\Rightarrow \angle BAF=\angle ADE=2x-\alpha,AF=DE,$又$\frac{DE}{DC}=\frac{BE}{BD}=\frac{\sqrt2}2,\frac{AE}{AB}=\frac{AC}{BC}=\frac{\sqrt2}2\Rightarrow DE=\frac{\sqrt2}2DC,AE=\frac{\sqrt2}2AB=\frac{\sqrt2}2AD\Rightarrow DE+AE=\frac{\sqrt2}2(DC+AD)=\frac{\sqrt2}2AC=\frac12BC=BE=BF+EF\Rightarrow EF=DE=AF\Rightarrow \angle EAF=\angle FEA=2x\Rightarrow \angle BAD=5x,$又$\angle ABD+\angle BDA+\angle BAD=180^\circ\Rightarrow 7x=180^\circ\Rightarrow \angle BAD=5x=\frac{900^\circ}7$