G4.1 In figure $1, A B C$ and $E B C$ are two right-angled triangles, $\angle B A C=\angle B E C=90^{\circ}, A B=A C$ and $E D B$ is the angle bisector of $\angle A B C$. Find the value of $\frac{B D}{C E}$.
$\triangle A B C$ is a right-angled isosceles triangle.
$\angle A B C=\angle A C B=45^{\circ}(\angle$ s sum of isos. $\triangle)$
$\angle A B D=\angle C B D=22.5^{\circ}$ ( $\angle$ bisector $)$
Let $B C=x$
$A B=x \cos 45^{\circ}=\frac{\sqrt{2} x}{2} ; C E=x \sin 22.5^{\circ}$ $B D=A B \div \cos 22.5^{\circ}=\frac{\sqrt{2} x}{2 \cos 22.5^{\circ}}$ $\frac{B D}{C E}=\frac{\sqrt{2} x}{2 \cos 22.5^{\circ} \cdot x \sin 22.5^{\circ}}=\frac{\sqrt{2}}{\cos 45^{\circ}}=2$
Method 2 Produce $C E$ and $B A$ to meet at $F$.
$A B=A C \Rightarrow \angle A B C=\angle A C B=45^{\circ}$
$\angle B A C=\angle B E C=90^{\circ}$ (given)
$\Rightarrow A B C E$ is a cyclic quad. (converse, $\angle$ s in the same seg.)
$\angle A B D=\angle C B D=22.5^{\circ}$ ( $\angle$ bisector $)$
$\angle A C F=22.5^{\circ}(\angle \mathrm{s}$ in the same seg. $)$
$\angle C A E=\angle C B E=22.5^{\circ}(\angle \mathrm{s}$ in the same seg. $)$
$C E=A E \ldots$ (1) (sides, opp. eq. $\angle$ s)
$\angle E A F=180^{\circ}-90^{\circ}-22.5^{\circ}=67.5^{\circ}$ (adj. $\angle \mathrm{s}$ on st. line)
$\angle A F E=180^{\circ}-90^{\circ}-22.5^{\circ}=67.5^{\circ}(\angle \mathrm{s}$ of $\triangle A C F)$
$A E=E F \ldots$ (2) (sides, opp. eq. $\angle$ s)
By (1) and (2), $C F=2 C E \ldots$ (3)
$\triangle A C F \cong \triangle A B D$ (A.S.A.)
$B D=C F$ (corr. sides, $\cong \Delta ' s$ s)
$\frac{B D}{C E}=\frac{C F}{C E}=\frac{2 C E}{C E}=2$ by (3)