《2025年1课3练江苏人民出版社八年级数学下册北师大版》

5 在△ABC中，∠ACB=2∠B，BC=2AC. 求证：∠A=90°.

答案：
如图，作$\angle ACB$的平分线$CD$交$AB$于点$D$，过点$D$作$DE\perp BC$于点$E$.

$\therefore \angle ACD = \angle ECD = \frac{1}{2}\angle ACB.$
$\because \angle ACB = 2\angle B,\therefore \angle B = \angle ECD,$
$\therefore BD = CD.\because DE\perp BC$于点$E,\therefore BE = CE = \frac{1}{2}BC.$
$\because BC = 2AC,\therefore AC = EC.$
在$\triangle ACD$和$\triangle ECD$中，$\begin{cases}AC = EC, \\\angle ACD = \angle ECD, \\CD = CD,\end{cases}$
$\therefore \triangle ACD\cong\triangle ECD(SAS),\therefore \angle A = \angle CED = 90^{\circ}.$

6 如图，在△ABC中，AD⊥BC于点D，∠B=2∠C. 求证：AB+BD=CD.
　　　　　　　　　　

答案：
如图，在$CD$上取一点$E$，使$ED = BD$，连接$AE$.

$\because AD\perp BC,\therefore \angle ADE = \angle ADB = 90^{\circ}.$
在$\triangle ADE$和$\triangle ADB$中，$\begin{cases}AD = AD, \\\angle ADE = \angle ADB, \\ED = BD,\end{cases}$
$\therefore \triangle ADE\cong\triangle ADB,$
$\therefore AE = AB,\angle AEB = \angle B = 2\angle C.$
又$\angle AEB = \angle C + \angle EAC,$
$\therefore \angle EAC = \angle C,\therefore AE = EC,\therefore AB = EC,$
$\therefore CD = EC + ED = AB + BD$，即$AB + BD = CD.$

7 如图，已知在△ABC中，AB=AC，D是△ABC外一点，且∠ABD=60°，∠ACD=60°. 求证：BD+DC=AB.

答案：
如图，延长$BD$到点$F$，使$BF = BA$，连接$AF$，$CF$.
$\because \angle ABD = 60^{\circ},BF = BA,$
$\therefore \triangle ABF$为等边三角形，
$\therefore AF = AB = AC = BF,\angle AFB = 60^{\circ},$

$\therefore \angle ACF = \angle AFC.$
又$\angle ACD = 60^{\circ},$
$\therefore \angle AFB = \angle ACD = 60^{\circ},$
$\therefore \angle DFC = \angle DCF,\therefore DC = DF,$
$\therefore BD + DC = BD + DF = BF = AB$，即$BD + DC = AB.$

8 如图，在四边形ABCD中，AD=4，BC=1，∠A=30°，∠B=90°，∠ADC=120°. 求CD的长.
　　　　　　　　　　　

答案：
如图，延长$AD$，$BC$交于点$E$.

$\because \angle A = 30^{\circ},\angle B = 90^{\circ},\therefore BE = \frac{1}{2}AE,\angle E = 60^{\circ}.$
$\because \angle ADC = 120^{\circ},\therefore \angle EDC = 60^{\circ},\therefore \angle E = \angle EDC = \angle ECD = 60^{\circ},\therefore \triangle EDC$是等边三角形.
设$CD = CE = DE = x.\because AD = 4,BC = 1,$
$\therefore 1 + x = \frac{1}{2}(x + 4)$，解得$x = 2.\therefore CD = 2.$