닮음과 둘레

두 삼각형 $\mathrm{ABC}$, $\mathrm{DBA}$에서 $\mathrm{\angle} BAD = \angle BCA$, $\mathrm{\angle} B$는 공통이므로 $\mathrm{\triangle} ABC \backsim \triangle DBA$ $\displaystyle \mathrm{\overline{CD}} = \mathit{x}$, $\displaystyle \mathrm{\overline{AC}} = \mathit{x} - 1$, $\displaystyle \mathrm{\overline{AB}} = \mathit{y}$라 하면 $\displaystyle \mathrm{\overline{AB}} : {\overline{AC}} = {\overline{DB}} : {\overline{DA}}$이므로 $y : \left( x - 1 \right) = 8 : 6$ $\therefore$ $\displaystyle x = \frac{3}{4} y + 1 \cdots \text{㉠}$ $\displaystyle \mathrm{\overline{AB}} : {\overline{BC}} = {\overline{DB}} : {\overline{BA}}$이므로 $y : \left( 8 + x \right) = 8 : y$ $\therefore$ $y ^{2} = 8 x + 64 \cdots \text{㉡}$ $\text{㉠}$을 $\text{㉡}$에 대입하면 $\displaystyle y ^{2} = 8 \times \left( \frac{3}{4} y + 1 \right) + 64$ $y ^{2} - 6 y - 72 = 0$, $\left( y - 12 \right) \left( y + 6 \right) = 0$ $\therefore$ $y = 12$ $\left( \because y > 0 \right)$ $y = 12$를 $\text{㉠}$에 대입하면 $x = 10$ $\therefore$ $\displaystyle \mathrm{\overline{AB}} = 12$, $\displaystyle \mathrm{\overline{BC}} = 8 + 10 = 18$, $\displaystyle \mathrm{\overline{CA}} = 9$ 따라서 삼각형 $\mathrm{ABC}$의 둘레의 길이는 $12 + 18 + 9 = 39$