26. In the following diagram, $\angle A C B=90^{\circ}, D E \perp B C, B E=A C, B D=\frac{1}{2} \mathrm{~cm}$, and $D E+B C=1 \mathrm{~cm}$. Suppose $\angle A B C=x^{\circ}$. Find the value of $x$.
26. Ans: 30
Produce $B C$ to $F$ such that $C F=D E$. Then $\triangle B D E$ and $\triangle A C F$ are right-angled triangles. Note that $B E=A C$ and $D E=C F$. Thus, we have
$$
\triangle B D E \cong \triangle A C F \text {. }
$$

Therefore, $A F=B D=\frac{1}{2}$, and $\angle F A C=\angle A B C$. Since $\angle B A C+\angle A B C=90^{\circ}$, we have
$$
\angle B A C+\angle F A C=90^{\circ} .
$$

Hence, $\angle B A F=90^{\circ}$. Now in $\triangle B A F, A F=\frac{1}{2}$, and
$$
B F=B C+C F=B C+D E=1 .
$$

Hence,
$$
\sin x^{\circ}=\frac{A F}{B F}=\frac{1}{2} \Longrightarrow x=30 .
$$

Alternative solution: Let $B E=a$ and $D E=b$. Since $\angle B E D=90^{\circ}$, it follows from Pythagoras Theorem that
$$
a^{2}+b^{2}=\frac{1}{4} .
$$

Since $D E+B C=1$, we have $B C=1-b$. As $\triangle B D E \sim \triangle A C F$, we have
$$
\frac{b}{a}=\frac{a}{1-b} \Longrightarrow a^{2}=b(1-b) .
$$

Together with the previous equality, we have
$$
b(1-b)+b^{2}=\frac{1}{4} \Longrightarrow b=\frac{1}{4}
$$

Hence,
$$
\sin x^{\circ}=\frac{D E}{B D}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2} \Longrightarrow x=30 .
$$