《2025年名校题库八年级数学上册北师大版》

14.(武侯区期末)在四边形ABCD中,$\angle BAD= 90^{\circ}$,$AB= AD$.
(1)如图1,若$AB= 2$,$BC= \sqrt{2}$,$CD= \sqrt{6}$.
①连接BD,试判断$\triangle BCD$的形状,并说明理由;
②连接AC,过点A作$AE\perp AC$,交CD的延长线于点E,求$\triangle ACE$的面积.
(2)如图2,若$\angle BCD= 135^{\circ}$,$BC= 2\sqrt{5}$,四边形ABCD的面积为$\frac{35}{2}$,求CD的长.

答案：
解:
(1)①$△BCD$是直角三角形。理由如下:$\because ∠BAD = 90^{\circ },AB = AD = 2,\therefore DB = \sqrt{2}AB = 2\sqrt{2}$。$\because DB^{2} = 8,BC^{2} + CD^{2} = 2 + 6 = 8,\therefore DB^{2} = DC^{2} + BC^{2}$，$\therefore ∠DCB = 90^{\circ },\therefore △BCD$是直角三角形。
②$\because ∠DAB = ∠DCB = 90^{\circ },\therefore ∠ADC + ∠ABC = 180^{\circ }$。$\because ∠ADE + ∠ADC = 180^{\circ },\therefore ∠ADE = ∠ABC$。$\because AE⊥AC,\therefore ∠EAC = ∠DAB = 90^{\circ },\therefore ∠DAE = ∠BAC$。又$\because AB = AD,\therefore △ABC≌△ADE(ASA)$，$\therefore AE = AC,DE = BC = \sqrt{2}$，$\therefore EC = DE + DC = \sqrt{2} + \sqrt{6}$。$\because AE = AC,AE⊥AC,\therefore EC = \sqrt{2}AE,\therefore AE = AC = \sqrt{3} + 1$，$\therefore S_{△AEC} = \frac{1}{2}AE\cdot AC = 2 + \sqrt{3}$。
(2)如图，过点B作$BM⊥DC$，交DC的延长线于点M，连接AM，过点A作$AP⊥AM$，交MD的延长线于点P。$\because BM⊥CD,\therefore ∠BMC = 90^{\circ }$。$\because ∠BCD = 135^{\circ },\therefore ∠BCM = 45^{\circ },\therefore ∠BCM = ∠CBM = 45^{\circ },\therefore CM = BM$。$\because BC^{2} = CM^{2} + BM^{2} = 20,\therefore BM = CM = \sqrt{10}$。$\because ∠DAB = ∠DMB = 90^{\circ },\therefore ∠ADC + ∠ABM = 180^{\circ }$。$\because ∠ADP + ∠ADC = 180^{\circ },\therefore ∠ADP = ∠ABM$。$\because AP⊥AM,\therefore ∠PAM = ∠DAB = 90^{\circ },\therefore ∠DAP = ∠BAM$。又$\because AB = AD,\therefore △ABM≌△ADP(ASA)$，$\therefore AP = AM,DP = BM = \sqrt{10},S_{△APD} = S_{△ABM}$，$\therefore S_{△APM} = S_{四边形ABMD}$。$\because S_{四边形ABCD} = S_{四边形ABMD} - S_{△BCM},\therefore \frac{35}{2} = S_{△APM} - \frac{1}{2}×(\sqrt{10})^{2}$，$\therefore S_{△APM} = \frac{45}{2}$，即$\frac{1}{2}\cdot AP^{2} = \frac{45}{2}$，$\therefore AP^{2} = 45$。$\because AP^{2} + AM^{2} = PM^{2},\therefore PM^{2} = 45 + 45 = 90$，$\therefore PM = 3\sqrt{10}$，$\therefore CD = PM - PD - CM = \sqrt{10}$。