20.(9分)如图7-4-12,在$△ ABC$中,点D在BC边上,$∠ B = ∠4,∠1 = ∠2 = ∠3$.求证:$BC = DE$.

20. 证明: $\because ∠B = ∠4$, $\therefore AB = AD$.
$\because ∠1 = ∠2 = ∠3$,
$\therefore ∠1 + ∠CAD = ∠2 + ∠CAD$,
即 $∠BAC = ∠DAE$.
又 $\because ∠1 + ∠4 + ∠B = ∠ADE + ∠3 + ∠4$,
$\therefore ∠B = ∠ADE$.
在$△ABC$ 和 $△ADE$ 中,
$\begin{cases}∠BAC = ∠DAE, \\AB = AD, \\∠B = ∠ADE,\end{cases}$
$\therefore △ABC ≌ △ADE (\mathrm{ASA})$.
$\therefore BC = DE$.

21.(9分)如图7-4-13,在△ABC中,∠B=90°,AB=16cm,BC=12cm,P,Q是△ABC边上的两个动点,其中点P从点A开始沿A→B路径运动,速度为1cm/s,点Q从点B开始沿B→C→A路径运动,速度为2cm/s,点P,Q同时出发,设出发后经过的时间为t s.
(1)斜边AC的长是cm,点P,Q出发2s后,PQ的长是cm.
(2)当点Q在BC边上运动时,出发几秒后,△PQB是等腰三角形?
(3)当点Q在CA边上运动时,求能使△BCQ成为等腰三角形的运动时间.(求出两种情况即可)

21. 解:(1) 20; $2\sqrt{53}$(或$\sqrt{212}$).
(2) 由题意得 $BQ = 2t$,$BP = (16 - t)$,
当 $BQ = BP$ 时, $2t = 16 - t$. 解得 $t = \frac{16}{3}$.
即出发 $\frac{16}{3}$ s 后,$△PQB$ 是等腰三角形.
(3) ① 当 $CQ = BC$ 时,$BC + CQ = 24 (\mathrm{cm})$.
$\therefore t = 24 ÷ 2 = 12 (\mathrm{s})$.
② 当 $CQ = BQ$ 时,$∠C = ∠CBQ$.
$\because ∠CBQ + ∠ABQ = ∠ABC = 90°$,
又 $\because ∠A + ∠C = 90°$.
$\therefore ∠A = ∠ABQ$.
$\therefore BQ = AQ$.
$\therefore CQ = AQ = 10 (\mathrm{cm})$.
$\therefore BC + CQ = 22 (\mathrm{cm})$.
$\therefore t = 22 ÷ 2 = 11 (\mathrm{s})$.
③ 当 $BC = BQ$ 时,过点 $B$ 作 $BD ⊥ AC$ 于点 $D$.
由三角形面积得 $BD = \frac{AB · BC}{AC} = \frac{48}{5} (\mathrm{cm})$.
由勾股定理得 $CD^2 = BC^2 - BD^2$,
即 $CD = \frac{36}{5} = 7.2 (\mathrm{cm})$.
$\therefore CQ = 2CD = 14.4 (\mathrm{cm})$.
$BC + CQ = 26.4 (\mathrm{cm})$.
$\therefore t = 26.4 ÷ 2 = 13.2 (\mathrm{s})$.
综上所述,当点 $P$,$Q$ 出发时间为 $11\mathrm{s}$,或 $12\mathrm{s}$,或 $13.2\mathrm{s}$ 时,$△BCQ$ 为等腰三角形.