Difficult Geometry - Angle #1

$\angle CAE=90^{\circ}-\angle CAD = 90^{\circ} - \angle BAD = \angle BAF \text{ } \cdots \text{ }\diamondsuit$

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$\angle FCE = \angle FBE = 90^{\circ}$

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$\angle BFE = 90^{\circ} - \angle FEB \text{ } \cdots \text{ } \spadesuit$

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Using $\spadesuit$ , we can get

$\angle ACE = \angle FCE - \angle FCA = 90^{\circ} - \angle FCA = 90^{\circ} - \angle FCB = 90^{\circ} - \angle FEB = \angle BFE \text{ } \cdots \text{ } \heartsuit$

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Since $\diamondsuit$ and $\heartsuit$ are both true, we can say that

$\triangle ACE \sim \triangle AFB \quad \text{(AA similarity)}$

$\therefore \text{ } AC:AE=AF:AB \text{, } AE\times AF = AB\times AC$ .

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Therefore, according to the power theorem ,

$AD^2=AE\times AF = AB\times AC = 3 \times 6 = 18$ .

$AD=3\sqrt{2}$

$\therefore\text{ } AD^3=AD^2\times AD=18 \times 3\sqrt{2}=\boxed{54\sqrt{2}}=a\sqrt{b}$

$a=54\text{, }b=2$

$\therefore\text{ }\boxed{a+b=56}$

Extend BA to meet the circle in C', Now, <CAE=<BAF as shown in the solution already presented.and <BAF=<EAC' and thus AC=AC'. But we know that BAxAC' = AExAF or BAxAC=AExAF or AExAF=6x3=18=AD². Hence etc.