1. 如图,在$△ ABC$中,$AD⊥ BC$于点$D$,给出下列等积式:①$AD^2=BD· CD$;②$AB· CD=AC· AD$;③$AC^2=BC· CD$;④$AB^2=AC· BD$.其中能证明$∠ BAC=90°$的有 （）

A.①②④
B.①③④
C.①②③
D.②③④

1. C 【点拨】$\because AD ⊥ BC, \therefore ∠ ADC = ∠ ADB = 90°$.
$\because AD^2 = BD · CD, \therefore \frac{AD}{BD} = \frac{CD}{AD}. \therefore △ ADC ∽ △ BDA.$
$\therefore ∠ DAC = ∠ ABD. \therefore ∠ DAC + ∠ BAD = ∠ ABD + ∠ BAD = 90°,即∠ BAC = 90°,故①符合题意. \because AB · CD = AC · AD, \therefore \frac{AB}{AC} = \frac{AD}{CD}. 由∠ ADB = ∠ ADC = 90°,$
易得$△ ABD ∽ △ CAD, \therefore ∠ ABD = ∠ CAD. \therefore ∠ BAD + ∠ CAD = ∠ BAD + ∠ ABD = 90°, \therefore ∠ BAC = 90°,故②符合题意. \because AC^2 = BC · CD, \therefore \frac{AC}{BC} = \frac{CD}{AC}. 又\because ∠ ACD = ∠ BCA, \therefore △ ACD ∽ △ BCA. \therefore ∠ ADC = ∠ BAC = 90°,$
故③符合题意. 由$AB^2 = AC · BD$不能证明$∠ BAC = 90°$,故④不符合题意.

2. 如图，在$△ ABC$中，$∠ ACB=90°$，$P$为$△ ABC$内部一点且始终满足$∠ ACP=∠ CBP$，延长$BP$交$AC$于点$D$。
(1)求证：$CD^2=DP· DB$。
(2)若$∠ APB=90°+∠ CBA$，求证：$AD=CD$。

2. 【证明】(1)$\because ∠ ACP = ∠ CBP, ∠ CDP = ∠ BDC,$
$\therefore △ CDP ∽ △ BDC. \therefore \frac{CD}{BD} = \frac{DP}{CD}. \therefore CD^2 = DP · DB.$
(2)$\because ∠ APB = 90° + ∠ CBA,$
$\therefore ∠ APD = 180° - (90° + ∠ CBA) = 90° - ∠ CBA.$
$\because ∠ ACB = 90°, \therefore ∠ CAB = 90° - ∠ CBA.$
$\therefore ∠ DPA = ∠ DAB.$
又$\because ∠ ADP = ∠ BDA, \therefore △ ADP ∽ △ BDA.$
$\therefore \frac{AD}{DB} = \frac{DP}{AD}. \therefore AD^2 = DP · DB.$
由(1)可知,$CD^2 = DP · DB, \therefore AD = CD.$

3. 如图,已知点G是$△ ABC$的重心,分别延长线段BG,CG交边AC,AB于点E,D,若$BE=15$,则BG的长是 ()

A.5
B.7.5
C.9
D.10

3. D 【点拨】$\because$ 点 G 是$△ ABC$的重心,$\therefore BG = 2GE.$
又$\because BE = BG + GE = 15. \therefore 3GE = 15. \therefore GE = 5, \therefore BG = 10,故选 D.$

4. 如图，$△ ABC$的重心为$G$，$△ ABC$和$△ GBC$在$BC$边上的高之比为 （）

A.$1:2$
B.$1:3$
C.$2:1$
D.$3:1$

4. D 【点拨】如图,连结 AG 并延长，交 BC 于点 F，$\because G$ 是$△ ABC$的重心，$\therefore AF = 3FG. \because AE$ 是$△ ABC$的高，$DG$ 是$△ GBC$的高，$\therefore DG // AE.$
$\therefore △ FDG ∽ △ FEA. \therefore AE : DG = AF : FG = 3 : 1.$

5. 如图,在菱形ABCD中,E为AB的中点,连结DE交对角线AC于点F,若菱形ABCD的周长为40 cm,AC=16 cm,求△ADF与菱形ABCD的面积比.

5. 【解】如图,连结 BD,交 AC 于点 O.
$\because$ 四边形 ABCD 是菱形，
$\therefore O$ 是 AC,BD 的中点,且$AC ⊥ BD.$
$\because$ 菱形 ABCD 的周长为 40 cm,$AC = 16$ cm，
$\therefore AD = 10$ cm,$AO = 8$ cm.
在$\mathrm{Rt}△ AOD$中,由勾股定理得$OD = \sqrt{10^2 - 8^2} = 6(\mathrm{cm}),$
$\therefore BD = 12 \mathrm{ cm}. \therefore S_{\mathrm{菱形}ABCD} = \frac{1}{2}AC · BD = 96 \mathrm{ cm}^2.$
$\because O$ 为 BD 的中点,E 为 AB 的中点，
$\therefore$ 点 F 是$△ ABD$的重心.
$\therefore AF = \frac{2}{3}AO = \frac{2}{3} × 8 = \frac{16}{3}(\mathrm{cm}).$
$\therefore S_{△ ADF} = \frac{1}{2}AF · OD = 16 \mathrm{ cm}^2.$
$\therefore S_{△ ADF}:S_{\mathrm{菱形}ABCD} = 16:96 = 1:6,$
即$△ ADF$与菱形 ABCD 的面积比为$1:6.$

6. 如图,在矩形ABCD中,E为边CD上一点,且$AE⊥BD$.
(1)求证:$AD^2=DE· DC$;
(2)F为线段AE延长线上一点,且满足$EF=CF=\frac{1}{2}BD$,求证:$CE=AD$.

6. 【证明】(1)$\because$ 四边形 ABCD 为矩形，
$\therefore ∠ BAD = ∠ ADE = 90°, AB = DC.$
$\therefore ∠ ABD + ∠ ADB = 90°.$
$\because AE ⊥ BD, \therefore ∠ DAE + ∠ ADB = 90°.$
$\therefore ∠ DAE = ∠ ABD.$
又$\because ∠ ADE = ∠ BAD, \therefore △ ADE ∽ △ BAD.$
$\therefore \frac{AD}{BA} = \frac{DE}{AD}. \therefore AD^2 = DE · BA.$
又$\because AB = DC, \therefore AD^2 = DE · DC.$
(2)连结 AC 交 BD 于点 O.
易得$∠ ADB = ∠ AED.$
又$\because ∠ FEC = ∠ AED, \therefore ∠ ADO = ∠ FEC.$
$\because$ 四边形 ABCD 为矩形,$\therefore OA = OD = \frac{1}{2}BD.$
又$\because EF = CF = \frac{1}{2}BD, \therefore OA = OD = EF = CF.$
$\therefore ∠ ADO = ∠ OAD, ∠ FEC = ∠ FCE.$
$\because ∠ ADO = ∠ FEC,$
$\therefore ∠ ADO = ∠ OAD = ∠ FEC = ∠ FCE.$
在$△ ODA$和$△ FEC$中,$\begin{cases} ∠ ODA = ∠ FEC, \\ ∠ OAD = ∠ FCE, \\ OD = FE, \end{cases}$
$\therefore △ ODA ≌ △ FEC. \therefore AD = CE.$