7. 如图,已知$\triangle ABC$中,$BD=AD$,F是高AD和BE的交点,若$FD=4,AF=2$,则线段BC的长度为 (
A. 6
B. 8
C. 10
D. 12
C
)A. 6
B. 8
C. 10
D. 12
答案：C
8. 如图,四边形ABCD中,$AB=AD,AC=5,∠DAB=∠DCB=90^{\circ }$,则四边形ABCD的面积为 (
A. 15
B. 12.5
C. 14.5
D. 17
B
)A. 15
B. 12.5
C. 14.5
D. 17
答案：B
9. 如图,在$\triangle ABC$中,作$∠BAC$的平分线交BC于点P,$CM⊥AP$于点N.若$∠CAB=30^{\circ },$$∠B=55^{\circ }$,则$∠BPM$的度数为
40°
.答案：40°
10. 如图,$AB=AC,CD// AB$,点E是AC上一点,$∠ABE=∠CAD$,延长BE交AD于点F.
(1)求证:$\triangle ABE\cong \triangle CAD;$
证明：∵CD//AB，∴∠BAE = ∠ACD，
在△ABE和△CAD中，$\left\{\begin{array}{l} \angle BAE = \angle ACD, \\ AB = AC, \\ \angle ABE = \angle CAD, \end{array}\right.$
∴△ABE≌△CAD(
(2)如果$∠ABC=∠ACB=70^{\circ },∠ABE=25^{\circ }$,求$∠D$的度数.
解：∵∠ACB = ∠ABC = 70°，∴∠BAC = 180° - ∠ABC - ∠ACB = 180° - 70° - 70° = 40°。
∵∠ABE = 25°，∴∠AEB = 180° - ∠ABE - ∠BAE = 180° - 25° - 40° = 115°。
由(1)知△ABE≌△CAD，∴∠D = ∠AEB =
(1)求证:$\triangle ABE\cong \triangle CAD;$
证明：∵CD//AB，∴∠BAE = ∠ACD，
在△ABE和△CAD中，$\left\{\begin{array}{l} \angle BAE = \angle ACD, \\ AB = AC, \\ \angle ABE = \angle CAD, \end{array}\right.$
∴△ABE≌△CAD(
ASA
)。(2)如果$∠ABC=∠ACB=70^{\circ },∠ABE=25^{\circ }$,求$∠D$的度数.
解：∵∠ACB = ∠ABC = 70°，∴∠BAC = 180° - ∠ABC - ∠ACB = 180° - 70° - 70° = 40°。
∵∠ABE = 25°，∴∠AEB = 180° - ∠ABE - ∠BAE = 180° - 25° - 40° = 115°。
由(1)知△ABE≌△CAD，∴∠D = ∠AEB =
115°
。答案：(1)证明：∵CD//AB，∴∠BAE = ∠ACD，
在△ABE和△CAD中，$\left\{\begin{array}{l} \angle BAE = \angle ACD, \\ AB = AC, \\ \angle ABE = \angle CAD, \end{array}\right.$
∴△ABE≌△CAD(ASA)。
(2)解：∵∠ACB = ∠ABC = 70°，∴∠BAC = 180° - ∠ABC - ∠ACB = 180° - 70° - 70° = 40°。
∵∠ABE = 25°，∴∠AEB = 180° - ∠ABE - ∠BAE = 180° - 25° - 40° = 115°。
由(1)知△ABE≌△CAD，∴∠D = ∠AEB = 115°。
在△ABE和△CAD中，$\left\{\begin{array}{l} \angle BAE = \angle ACD, \\ AB = AC, \\ \angle ABE = \angle CAD, \end{array}\right.$
∴△ABE≌△CAD(ASA)。
(2)解：∵∠ACB = ∠ABC = 70°，∴∠BAC = 180° - ∠ABC - ∠ACB = 180° - 70° - 70° = 40°。
∵∠ABE = 25°，∴∠AEB = 180° - ∠ABE - ∠BAE = 180° - 25° - 40° = 115°。
由(1)知△ABE≌△CAD，∴∠D = ∠AEB = 115°。
11. (1)如图①,$\triangle ABC$中,$AB=AC,∠BAC=45^{\circ },CD⊥AB,AE⊥BC$,垂足分别为D,E,CD与AE交于点F.
①写出图①中所有的全等三角形:____;
②线段AF与线段CE的数量关系是:____.
(2)问题探究:如图②,$\triangle ABC$中,$AB=BC,∠BAC=∠BCA=$$45^{\circ }$,AD平分$∠BAC,AD⊥CD$,垂足为D,AD与BC交于点E. 求证:$AE=2CD.$
①写出图①中所有的全等三角形:____;
②线段AF与线段CE的数量关系是:____.
(2)问题探究:如图②,$\triangle ABC$中,$AB=BC,∠BAC=∠BCA=$$45^{\circ }$,AD平分$∠BAC,AD⊥CD$,垂足为D,AD与BC交于点E. 求证:$AE=2CD.$
答案：
(1)①△ABE≌△ACE，△ADF≌△CDB
②AF = 2CE
(2)证明：延长AB，CD交于点G，如答图。
∵AD平分∠BAC，∴∠CAD = ∠GAD，∵AD⊥CD，∴∠ADC = ∠ADG = 90°。
在△ADC和△ADG中，$\left\{\begin{array}{l} \angle ADC = \angle ADG, \\ AD = AD, \\ \angle CAD = \angle GAD, \end{array}\right.$
∴△ADC≌△ADG(ASA)，∴CD = GD，即CG = 2CD，
∵∠BAC = 45°，AB = BC，∴∠ABC = 90°，
∴∠CBG = 90°，∴∠G + ∠BCG = 90°，
∵∠G + ∠BAE = 90°，∴∠BAE = ∠BCG，
在△ABE和△CBG中，$\left\{\begin{array}{l} \angle ABE = \angle CBG, \\ AB = CB, \\ \angle BAE = \angle BCG, \end{array}\right.$
∴△ABE≌△CBG(ASA)，∴AE = CG = 2CD。
(1)①△ABE≌△ACE，△ADF≌△CDB
②AF = 2CE
(2)证明：延长AB，CD交于点G，如答图。
∵AD平分∠BAC，∴∠CAD = ∠GAD，∵AD⊥CD，∴∠ADC = ∠ADG = 90°。
在△ADC和△ADG中，$\left\{\begin{array}{l} \angle ADC = \angle ADG, \\ AD = AD, \\ \angle CAD = \angle GAD, \end{array}\right.$
∴△ADC≌△ADG(ASA)，∴CD = GD，即CG = 2CD，
∵∠BAC = 45°，AB = BC，∴∠ABC = 90°，
∴∠CBG = 90°，∴∠G + ∠BCG = 90°，
∵∠G + ∠BAE = 90°，∴∠BAE = ∠BCG，
在△ABE和△CBG中，$\left\{\begin{array}{l} \angle ABE = \angle CBG, \\ AB = CB, \\ \angle BAE = \angle BCG, \end{array}\right.$
∴△ABE≌△CBG(ASA)，∴AE = CG = 2CD。