【例题1】如图,过半径为$6 cm的\odot O外一点P引圆的切线PA,PB$,连接$PO交\odot O于点F$,过点$F作\odot O的切线分别交PA,PB于点D,E$. 若$PO为10 cm$,$\angle APB = 40^{\circ}$.
(1)求$\triangle PED$的周长.
(2)求$\angle DOE$的度数.

思路导引 由切线长定理知,$PA = PB$,$DA = DF$,所以$\triangle PED$的周长可转化为两条切线长之和. 在有关圆的计算中,经常作过切点的半径构建直角三角形,从而利用勾股定理计算.
解：(1)如图,连接$AO,BO$,则$OA \perp PA$,那么$PA = \sqrt{PO^{2} - AO^{2}} = \sqrt{10^{2} - 6^{2}} = 8(cm)$.

$\because PA,PB,DE均为\odot O$的切线,$A,B,F$为切点,$\therefore PA = PB$,$DF = DA$,$EF = EB$.
$\therefore \triangle PED的周长为PD + DF + EF + PE = PA + PB = 8 + 8 = 16(cm)$.
(2)由切线长定理,得$\angle ADO = \angle FDO$,$\angle OEB = \angle OEF$,$\therefore \angle 1 = \angle 2 = \frac{1}{2}\angle AOF$,$\angle FOE = \angle BOE = \frac{1}{2}\angle BOF$. $\therefore \angle DOE = \angle 2 + \angle FOE = \frac{1}{2}\angle AOB$.
$\because \angle AOB + \angle APB = 180^{\circ}$,$\therefore \angle DOE = \frac{1}{2}(180^{\circ} - \angle APB) = \frac{1}{2}(180^{\circ} - 40^{\circ}) = 70^{\circ}$.

【例题2】$\odot O分别切\triangle ABC的三边AB,BC,CA于点D,E,F$,设$BC = a,AC = b,AB = c$.
(1)如图①,求$AD,BE,CF$的长.
(2)如图②,当$\angle ACB = 90^{\circ}$时,求内切圆的半径的长.

思路导引 (1)小题可运用切线长的性质列方程组求解. (2)小题是(1)小题的特殊情形,可以结合三角形的面积来求解.
解：(1)如图①,设$AD = x,BE = y,CF = z$. 由切线长的性质可知,$AD = AF,BD = BE,CE = CF$,则有$\begin{cases}x + y = c, \\y + z = a, \\z + x = b.\end{cases} 解得\begin{cases}x = \frac{b + c - a}{2}, \\y = \frac{a + c - b}{2}, \\z = \frac{a + b - c}{2}.\end{cases} $
即$AD = \frac{b + c - a}{2}$,$BE = \frac{a + c - b}{2}$,$CF = \frac{a + b - c}{2}$.
(2)如图②,连接$OD,OE,OF$,则$OD \perp AB$,$OF \perp AC$,$OE \perp BC$.

又$\angle ACB = 90^{\circ}$,$OF = OE$,$\therefore四边形OFCE$是正方形.
$\therefore CF = CE = r$.
$\therefore AD = AF = b - r$,$BD = BE = a - r$.
$\because AD + BD = c$,
$\therefore b - r + a - r = c$. $\therefore r = \frac{a + b - c}{2}$.
即直角三角形内切圆的半径等于两直角边的和与斜边的差的一半.
结合三角形的面积求解：
连接$OA,OB,OC,OD,OE,OF$.
$\because S_{\triangle AOB} = \frac{1}{2}AB \cdot OD = \frac{1}{2}c \cdot r$,$S_{\triangle AOC} = \frac{1}{2}AC \cdot OF = \frac{1}{2}b \cdot r$,$S_{\triangle BOC} = \frac{1}{2}BC \cdot OE = \frac{1}{2}a \cdot r$,$\therefore S_{\triangle ABC} = S_{\triangle AOB} + S_{\triangle AOC} + S_{\triangle BOC} = \frac{1}{2}r(a + b + c)$.
又$S_{\triangle ABC} = \frac{1}{2}AC \cdot BC = \frac{1}{2}ab$,
$\therefore \frac{1}{2}r(a + b + c) = \frac{1}{2}ab$. 即$r = \frac{ab}{a + b + c}$.
故直角三角形内切圆的半径等于两直角边的积除以周长所得的商.
(本例题中的相关结论在今后的计算中十分有用.)