G4.4 In figure $1, B E C$ is a semicircle and $F$ is a point on the diameter $B C$. Given that $B F: F C=3: 1, A B=8$ and $A E=4$. Find the length of $E C$.
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Join $B E$. It is easy to show that $\triangle B E F \sim \triangle E C F$ (equiangular)
Let $B F=3 k, C F=k$
$E F: 3 k=k: E F \quad$ (ratio of sides, $\sim \Delta$ 's)
$E F=\sqrt{3} k$
$B E^{2}=B F^{2}+E F^{2}=9 k^{2}+3 k^{2}$ (Pythagoras' Theorem on $\triangle B E F$ ) $\Rightarrow B E=\sqrt{12} k$
$\angle B E C=90^{\circ} \quad$ ( $\angle$ in semi-circle)
$B E^{2}+A E^{2}=A B^{2} \quad$ (Pythagoras' Theorem on $\triangle A B E$ )
$12 k^{2}+16=64 \Rightarrow k=2$
$E C^{2}=C F^{2}+E F^{2}=2^{2}+3 \times 2^{2}$ (Pythagoras' Theorem on $\triangle C E F$ )
$E C=4$
Method 2
$\angle B E C=90^{\circ}$ ( $\angle$ in semi-circle $)$
$\angle B E A=90^{\circ}$ (adj. $\angle$ s on st. line)
$\cos \angle B A E=\frac{A E}{A B}=\frac{4}{8}=\frac{1}{2}$
$\angle B A E=60^{\circ}$
Let $B F=3 k, C F=k, \angle E C B=\theta$.
$\angle C E F=90^{\circ}-\theta \quad(\angle \mathrm{s}$ sum of $\triangle C E F)$
$\angle C B E=90^{\circ}-\theta \quad(\angle \mathrm{s}$ sum of $\triangle B C E)$
$\angle B E F=\theta \quad(\angle \mathrm{s} \mathrm{sum}$ of $\triangle B E F)$
$\triangle C E F \sim \triangle E B F \quad$ (equiangular)
$\frac{C F}{E F}=\frac{E F}{B F} \quad$ (corr. sides, $\sim \Delta \mathrm{s}$ )
$E F^{2}=k \cdot 3 k$
$E F=\sqrt{3} k$
$\tan \angle E C F=\frac{E F}{C F}=\frac{\sqrt{3} k}{k}=\sqrt{3}$
$\angle E C F=60^{\circ}$
$\triangle A B E \cong \triangle C B E$
$E C=A E=4$
(A.A.S.)
(corr. sides, $\cong \Delta \mathrm{s}$ )
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