典例4 如图，在Rt△ABC中，∠ACB = 90°，∠B = 30°，BC = 8，AD是∠BAC的平分线. 若P，Q分别是AD和AC上的动点，求PC + PQ的最小值.
　　　　　　　　　　　　　

典例4如图,在AB上取一点Q',使得AQ' = AQ,连接PQ',CQ'.∵AD是∠BAC的平分线,∴∠PAQ = ∠PAQ'.∵AP = AP,∴△PAQ≌△PAQ'.∴PQ = PQ'.∴PC + PQ = PC + PQ'.∴当点C,P,Q'在同一条直线上,且CQ'⊥AB时,PC + PQ有最小值,最小值为CQ'的长,此时∠BQ'C = 90°.∵∠B = 30°,BC = 8,∴CQ' = $\frac{1}{2}$BC = 4.∴PC + PQ的最小值为4.
　　

典例5 如图，在△ABC中，∠ACB = 90°，AB + BC = 8，tan A = $\frac{3}{4}$，O，D分别是边AB，AC上的动点，求OC + OD的最小值.
　　　　　　　　　　　　　

典例5如图,作点C关于AB的对称点C',连接CC',交AB于点E,过点C'作C'D⊥AC于点D,交AB于点O,连接OC,则OC = OC'.易知此时OC + OD的值最小,即为C'D的长.在△ABC中,∠ACB = 90°,tanA = $\frac{3}{4}$,∴$\frac{BC}{AC}=\frac{3}{4}$.设BC = 3x,则AC = 4x.∴AB = $\sqrt{BC^{2}+AC^{2}} = 5x$.∵AB + BC = 8,∴5x + 3x = 8,解得x = 1.∴BC = 3,AC = 4,AB = 5.由对称,可知CC'⊥AB,CE = C'E = $\frac{1}{2}$CC',∴S△ABC=$\frac{1}{2}$BC·AC=$\frac{1}{2}$AB·CE,即$\frac{1}{2}$×3×4 = $\frac{1}{2}$×5CE,解得CE = $\frac{12}{5}$.∴CC' = 2CE = $\frac{24}{5}$.∵CC'⊥AB,C'D⊥AC,∴∠C'EO = ∠ODA = ∠C'DC = 90°.∵∠C'OE = ∠AOD,∴∠C' = ∠A.又∵∠C'DC = ∠ACB = 90°,∴△C'DC∽△ACB.∴$\frac{C'D}{AC}=\frac{C'C}{AB}$,即$\frac{C'D}{4}=\frac{\frac{24}{5}}{5}$.∴C'D = $\frac{96}{25}$,即OC + OD的最小值为$\frac{96}{25}$.