上海民办上宝中学数学三角形解答题同步单元检测(Word版 含答案)

上海民办上宝中学数学三角形解答题同步单元检测(Word版 含答案)
上海民办上宝中学数学三角形解答题同步单元检测(Word版 含答案)

上海民办上宝中学数学三角形解答题同步单元检测(Word 版 含答案)

一、八年级数学三角形解答题压轴题(难)

1.直线MN 与直线PQ 垂直相交于O ,点A 在直线PQ 上运动,点B 在直线MN 上运动. (1)如图1,已知AE 、BE 分别是∠BAO 和∠ABO 角的平分线,点A 、B 在运动的过程中,∠AEB 的大小是否会发生变化?若发生变化,请说明变化的情况;若不发生变化,试求出∠AEB 的大小.

(2)如图2,已知AB 不平行CD ,AD 、BC 分别是∠BAP 和∠ABM 的角平分线,又DE 、CE 分别是∠ADC 和∠BCD 的角平分线,点A 、B 在运动的过程中,∠CED 的大小是否会发生变化?若发生变化,请说明理由;若不发生变化,试求出其值.

(3)如图3,延长BA 至G ,已知∠BAO 、∠OAG 的角平分线与∠BOQ 的角平分线及延长线相交于E 、F ,在△AEF 中,如果有一个角是另一个角的3倍,试求∠ABO 的度数.

【答案】(1)135°;(2)67.5°;(3)60°, 45° 【解析】 【分析】

(1)根据直线MN 与直线PQ 垂直相交于O 可知∠AOB=90°,再由AE 、BE 分别是∠BAO 和∠ABO 的角平分线得出1BAE OAB 2∠=∠,1

ABE ABO 2

∠=∠,由三角形内角和定理即可得出结论;

(2)延长AD 、BC 交于点F ,根据直线MN 与直线PQ 垂直相交于O 可得出∠AOB=90°,进而得出OAB OBA 90∠+∠=? ,故PAB MBA 270∠+∠=?,再由AD 、BC 分别是∠BAP 和∠ABM 的角平分线,可知1BAD BAP 2∠=

∠,1

ABC ABM 2

∠=∠,由三角形内角和定理可知∠F=45°,再根据DE 、CE 分别是∠ADC 和∠BCD 的角平分线可知

CDE DCE 112.5∠+∠=?,进而得出结论;

(3))由∠BAO 与∠BOQ 的角平分线相交于E 可知

1EAO BAO 2∠=∠,1

EOQ BOQ 2

∠=∠ ,进而得出∠E 的度数,由AE 、AF 分别是∠BAO

和∠OAG 的角平分线可知∠EAF=90°,在△AEF 中,由一个角是另一个角的3倍分四种情况进行分类讨论. 【详解】

(1)∠AEB 的大小不变,

∵直线MN 与直线PQ 垂直相交于O ,

∴∠AOB=90°,

∴OAB OBA 90∠+∠=?,

∵AE 、BE 分别是∠BAO 和∠ABO 角的平分线, ∴1BAE OAB 2∠=

∠,1

ABE ABO 2

∠=∠, ∴()1

BAE ABE OAB ABO 452

∠+∠=∠+∠=°, ∴∠AEB=135°;

(2)∠CED 的大小不变. 如图2,延长AD 、BC 交于点F . ∵直线MN 与直线PQ 垂直相交于O , ∴90∠=AOB °,

∴OAB OBA 90∠+∠=°, ∴PAB MBA 270∠+∠=°,

∵AD 、BC 分别是∠BAP 和∠ABM 的角平分线, ∴1BAD BAP 2∠=

∠,1

ABC ABM 2

∠=∠, ∴()1

BAD ABC PAB ABM 1352

∠+∠=

∠+∠=°,F 45∠=°, ∴FDC FCD 135∠+∠=°, ∴CDA DCB 225∠+∠=°,

∵DE 、CE 分别是∠ADC 和∠BCD 的角平分线, ∴CDE DCB 112.5∠+∠=°, ∴E 67.5∠=°;

(3)∵∠BAO 与∠BOQ 的角平分线相交于E , ∴1EAO BAO 2∠=

∠,1

EOQ BOQ 2

∠=∠ , ∴()11

E EOQ EAO BOQ BAQ ABO 22

∠=∠-∠=

∠-∠=∠, ∵AE 、AF 分别是∠BAO 和∠OAG 的角平分线, ∴EAF 90∠=°. 在△AEF 中,

∵有一个角是另一个角的3倍,故有:

①EAF 3E ∠=∠,E 30∠=°,ABO 60∠=°; ②EAF 3F ∠=∠,E 60∠=°,ABO 120∠=°; ③EAF 3E ∠=∠,E 22.5∠=°,ABO 45∠=°; ④EAF 3F ∠=∠,E 67.5∠=°,ABO 135∠=°. ∴∠ABO 为60°或45°. 【点睛】

本题考查的是三角形内角和定理,熟知三角形内角和是180°是解答此题的关键.

2.在一个三角形中,如果一个角是另一个角的3倍,这样的三角形我们称之为“灵动三角形”.如,三个内角分别为120°,40°,20°的三角形是“灵动三角形”. 如图,∠MON =60°,在射线OM 上找一点A ,过点A 作AB ⊥OM 交ON 于点B ,以A 为端点作射线AD ,交线段OB 于点C (规定0°< ∠OAC < 90°).

(1)∠ABO 的度数为 °,△AOB (填“是”或“不是”灵动三角形); (2)若∠BAC =60°,求证:△AOC 为“灵动三角形”; (3)当△ABC 为“灵动三角形”时,求∠OAC 的度数.

【答案】(1)30°;(2)详见解析;(3)∠OAC=80°或52.5°或30°. 【解析】 【分析】

(1)根据垂直的定义、三角形内角和定理求出∠ABO 的度数,根据“智慧三角形”的概念判断;

(2)根据“智慧三角形”的概念证明即可;

(3)分点C 在线段OB 和线段OB 的延长线上两种情况,根据“智慧三角形”的定义计算. 【详解】

(1)答案为:30°;是; (2)∵AB ⊥OM ∴∠B AO =90° ∵∠BAC =60°

∴∠OAC =∠B AO-∠BAC=30° ∵∠MON =60°

∴∠ACO =180°-∠OAC-∠MON =90° ∴∠ACO =3∠OAC , ∴△AOC 为“灵动三角形”;

(3)设∠OAC= x°则∠BAC=90-x, ∠ACB=60+x ,∠ABC=30°

∵△ABC为“智慧三角形”,

Ⅰ、当∠ABC=3∠BAC时,°,

∴30=3(90-x),∴x=80

Ⅱ、当∠ABC=3∠ACB时,

∴30=3(60+x)∴x= -50 (舍去)

∴此种情况不存在,

Ⅲ、当∠BCA=3∠BAC时,

∴60+x=3(90-x),

∴x=52.5°,

Ⅳ、当∠BCA=3∠ABC时,

∴60+x=90°,

∴x=30°,

Ⅴ、当∠BAC=3∠ABC时,

∴90-x=90°,

∴x=0°(舍去)

Ⅵ、当∠BAC=3∠ACB时,

∴90-x=3(60+x),

∴x= -22.5(舍去),

∴此种情况不存在,

∴综上所述:∠OAC=80°或52.5°或30°。

【点睛】

考查的是三角形内角和定理、“智慧三角形”的概念,用分类讨论的思想解决问题是解本题的关键.

3.如图四边形ABCD中,AD∥BC,∠BCD=90°,∠BAD的平分线AG交BC于点G.

(1)求证:∠BAG=∠BGA;

(2)如图2,∠BCD的平分线CE交AD于点E,与射线GA相交于点F,∠B=50°.

①若点E在线段AD上,求∠AFC的度数;

②若点E在DA的延长线上,直接写出∠AFC的度数;

(3)如图3,点P在线段AG上,∠ABP=2∠PBG,CH∥AG,在直线AG上取一点M,使∠PBM=∠DCH,请直接写出∠ABM:∠PBM的值.

【答案】(1)证明见解析;(2)①20°;②160°;(3)1

3

7

3

【解析】

【分析】

(1)根据AD//BC可知∠GAD=∠BGA,由AG平分∠BAD可知∠BAG=∠GAD,即可得答案.(2)①根据CF平分∠BCD,∠BCD=90°,可求出∠GCF的度数,由AD//BC可求出∠AEF 和∠DAB的度数,根据三角形外角的性质求出∠AFC的度数即可;②根据三角形外角性质求出即可;(3)根据M点在BP的上面和下面两种情况讨论,分别求出∠PBM和∠ABM 的值即可.

【详解】

(1)∵AD∥BC,

∴∠GAD=∠BGA,

∵AG平分∠BAD,

∴∠BAG=∠GAD,

∴∠BAG=∠BGA;

(2)①∵CF平分∠BCD,∠BCD=90°,

∴∠GCF=45°,

∵AD∥BC,∠ABC=50°,

∴∠AEF=∠GCF=45°;∠DAB=180°﹣50°=130°,

∵AG平分∠BAD,

∴∠BAG=∠GAD=65°,

∴∠AFC=65°﹣45°=20°;

②如图:

∵∠AGB=65°,∠BCF=45°,

∴∠AFC=∠CGF+∠BCF=115°+45°=160°;

(3)有两种情况:

①当M在BC的下方时,如图:∵∠ABC=50°,∠ABP=2∠PBG,

∴∠ABP=(100

3

)°,∠PBG=(50

3

)°,

∵AG∥CH,

∴∠BCH=∠AGB=65°,∵∠BCD=90°,∴∠DCH=∠PBM=90°﹣65°=25°,

∴∠ABM=∠ABP+∠PBM=(100

3

+25)°=(

175

3

)°,

∴∠ABM:∠PBM=(175

3

)°:25°=7

3

②当M在BC的上方时,如图:

同理得:∠ABM=∠ABP﹣∠PBM=(100

3

﹣25)°=(25

3

)°,

∴∠ABM:∠PBM=(25

3

)°:25°=1

3

综上,∠ABM:∠PBM的值是1

3

7

3

【点睛】

本题考查平行线的性质和三角形外角性质,熟练掌握平行线性质是解题关键.

4.Rt△ABC中,∠C=90°,点D、E分别是△ABC边AC、BC上的点,点P是一动点.令

∠PDA=∠1,∠PEB=∠2,∠DPE=∠α.

(1)若点P在线段AB上,如图(1)所示,且∠α=50°,则∠1+∠2= °;

(2)若点P在边AB上运动,如图(2)所示,则∠α、∠1、∠2之间的关系为:;

(3)若点P运动到边AB的延长线上,如图(3)所示,则∠α、∠1、∠2之间有何关系?猜想并说明理由.

(4)若点P运动到△ABC形外,如图(4)所示,则∠α、∠1、∠2之间的关系为:.【答案】(1)140°;(2)∠1+∠2=90°+α;(3)∠1=90°+∠2+α,理由见解

析;(4)∠2=90°+∠1﹣α.

【解析】

试题分析:(1)根据四边形内角和定理以及邻补角的定义,得出∠1+∠2=∠C+∠α,进而得出即可;

(2)利用(1)中所求的结论得出∠α、∠1、∠2之间的关系即可;

(3)利用三角外角的性质,得出∠1=∠C+∠2+α=90°+∠2+α;

(4)利用三角形内角和定理以及邻补角的性质可得出∠α、∠1、∠2之间的关系.

试题分析:(1)∵∠1+∠2+∠CDP+∠CEP=360°,∠C+∠α+∠CDP+∠CEP=360°,∴∠1+∠2=∠C+∠α,

∵∠C=90°,∠α=50°,

∴∠1+∠2=140°,

故答案为140;

(2)由(1)得∠α+∠C=∠1+∠2,

∴∠1+∠2=90°+∠α.

故答案为∠1+∠2=90°+∠α.

(3)∠1=90°+∠2+∠α.理由如下:如图③,

设DP与BE的交点为M,

∵∠2+∠α=∠DME,∠DME+∠C=∠1,

∴∠1=∠C+∠2+∠α=90°+∠2+∠α.

(4)如图④,

设PE 与AC 的交点为F , ∵∠PFD =∠EFC ,

∴180°-∠PFD =180°-∠EFC , ∴∠α+180°-∠1=∠C +180°-∠2, ∴∠2=90°+∠1-∠α. 故答案为∠2=90°+∠1-∠α

点睛:本题考查了三角形内角和定理和外角的性质、对顶角相等的性质,熟练掌握三角形外角的性质是解决问题的关键.

5.ABC 中,AD 是BAC ∠的平分线,AE BC ⊥,垂足为E ,作CF//AD ,交直线AE 于点F.设B α∠=,ACB β∠=.

()1若B 30∠=,ACB 70∠=,依题意补全图1,并直接写出AFC ∠的度数; ()2如图2,若ACB ∠是钝角,求AFC ∠的度数(用含α,β的式子表示);

()3如图3,若B ACB ∠∠>,直接写出AFC ∠的度数(用含α,β的式子表示).

【答案】(1)补图见解析,AFC 20∠=;(2) ()1

AFC 180βα2

∠=-

-;(3) ()1

AFC αβ2

∠=

-. 【解析】 【分析】

(1)先根据三角形内角和定理求出∠BAC 和∠CAE ,根据角平分线定义求出∠CAD ,即可求出答案;

(2)先根据三角形内角和定理求出∠BAC ,根据角平分线定义求出∠BAD ,根据三角形外角性质求出∠ADC ,根据三角形内角和定理求出∠DAE ,根据平行线的性质求出即可; (3)求出∠DAE 度数,根据平行线的性质求出即可. 【详解】 解:()1如图1,

B 30∠=,ACB 70∠=,

BAC 180B ACB 80∠∠∠∴=--=,

AD 是BAC ∠的平分线,

1

CAD CAB 402

∠∠∴==,

AE BC ⊥,

AEC 90∠∴=,

ACB 70∠=,

EAC 180907020∠∴=--=,

DAE CAD CAE 402020∠∠∠∴=-=-=,

CF//AD ,

AFC DAE 20∠∠∴==;

()2如图2,

ABC 中,BAC B ACB 180∠∠∠++=, ()BAC 180B ACB ∠∠∠∴=-+. ()180αβ=-+,

AD 是BAC ∠的平分线,

()11

BAD BAC 90αβ22

∠∠∴==-+,

()()11

ADE B BAD α90αβ90βα22

∠∠∠∴=+=+-+=--,

AE BC ⊥,

DAE ADE 90∠∠∴+=,

()1

DAE 90ADE βα2

∠∠∴=-=

-, CF//AD ,

DAE AFC 180∠∠∴+=,

()1

AFC 180βα2

∠∴=-

-; ()3如图3,

ABC 中,BAC B ACB 180∠∠∠++=, ()BAC 180B ACB ∠∠∠∴=-+, ()180αβ=-+,

AD 是BAC ∠的平分线,

()11

CAD BAC 90αβ22

∠∠∴==-+,

AE BC ⊥,

AEC 90∠∴=, ACB β∠=,

EAC 18090β90β∠∴=--=-,

()

()()11

DAE CAE CAD 90β90αβαβ22

∠∠∠??∴=-=----=-????.

【点睛】

本题考查了三角形内角和定理、三角形角平分线定义、三角形的高、平行线的性质等,熟练掌握相关的性质与定理是解题的关键.

6.在△ABC 中,点D 、E 分别在边AC 、BC 上(不与点A 、B 、C 重合),点P 是直线AB 上的任意一点(不与点A 、B 重合).设∠PDA=x,∠PEB=y,∠DPE=m,∠C=n. (1)如图,当点P 在线段AB 上运动,且n=90°时 ①若PD∥BC,PE∥AC,则m=_____; ②若m=50°,求x+y 的值.

(2)当点P 在直线AB 上运动时,直接写出x 、y 、m 、n 之间的数量关系.

【答案】(1)①90°,②140°;(2)详见解析.

【解析】

分析:(1)①证明四边形DPEC为平行四边形可得结论;

②根据四边形内角和为360°,列等式求出x+y的值;

(2)根据P、D、E位置的不同,分五种情况:①y-x=m+n,如图2,点P在BA的延长线上时,根据三角形的内角和与外角定理列等式,化简后得出结论;

②x-y=m-n,如图3,点P在BA的延长线上时,根据三角形的内角和与外角定理列等式,化简后得出结论;

③x+y=m+n,如图4,点P在线段BA上时,根据四边形的内角和为360°列等式,化简后得出结论;

④x-y=m+n,如图5,同理得出结论;

⑤y-x=m-n,如图6,同理得出结论.

详解:(1)①如图1,

∵PD∥BC,PE∥AC,

∴四边形DPEC为平行四边形,

∴∠DPE=∠C,

∵∠DPE=m,∠C=n=90°,

∴m=90°;

②∵∠ADP=x,∠PEB=y,

∴∠CDP=180°-x,∠CEP=180°-y,

∵∠C+∠CDP+∠DPE+∠CEP=360°,

∠C=90°,∠DPE=50°,

∴90°+180°-x+50°+180°-y=360°,

∴x+y=140°;

(2)分五种情况:

①y﹣x=m+n,如图2,

理由是:

∵∠DFP=n+∠FEC,∠FEC=180°﹣y,

∴∠DFP=n+180°﹣y,

∵x+m+∠DFP=180°,

∴x+m+n+180°﹣y=180°,

∴y﹣x=m+n;

②x﹣y=m﹣n,如图3,

理由是:

同理得:m+180°﹣x=n+180°﹣y,

∴x﹣y=m﹣n;

③x+y=m+n,如图4,

理由是:

由四边形内角和为360°得:180°﹣x+m+180°﹣y+n=360°,∴x+y=m+n;

④x﹣y=m+n,如图5,

理由是:

同理得:180°=m+n+y+180°﹣x,

∴x﹣y=m+n;

⑤y﹣x=m﹣n,如图6,

理由是:

同理得:n+180°﹣x=m+180°﹣y,

∴y﹣x=m﹣n.

点睛:本题考查了三角形综合、平行四边形的判定.

7.如图①,在平面直角坐标系中,A(0,1),B(4,1),C为x轴正半轴上一点,且AC平分∠OAB.

(1)求证:∠OAC=∠OCA;

(2)如图②,若分别作∠AOC的三等分线及∠OCA的外角的三等分线交于点P,即满足∠POC

=1

3

∠AOC,∠PCE=

1

3

∠ACE,求∠P的大小;

(3)如图③,在(2)中,若射线OP、CP满足∠POC=1

n

∠AOC,∠PCE=

1

n

∠ACE,猜想∠OPC

的大小,并证明你的结论(用含n的式子表示).

【答案】(1)证明见解析(2)15°(3)45 n

【解析】

试题分析:(1)根据AB坐标可以求得∠OAB大小,根据角平分线性质可求得∠OAC大小,即可解题;

(2)根据题干中给出的∠POC=1

3

∠AOC、∠PCE=

1

3

∠ACE可以求得∠PCE和∠POC的大小,

再根据三角形外角等于不相邻两内角和即可解题;

(3)解法和(2)相同,根据题干中给出的∠POC=1

n

∠AOC、∠PCE=

1

n

∠ACE可以求得

∠PCE和∠POC的大小,再根据三角形外角等于不相邻两内角和即可解题.

试题解析:(1)证明:∵A(0,1),B(4,1),∴AB∥CO,∴∠OAB=180°-∠AOC=90°.∵AC平分∠OAB,∴∠OAC=45°,∴∠OCA=90°-45°=45°,∴∠OAC=∠OCA.

(2)解:∵∠POC=∠AOC,∴∠POC=×90°=30°.∵∠PCE=∠ACE,∴∠PCE=

(180°-45°)=45°.∵∠P+∠POC=∠PCE,∴∠P=∠PCE-∠POC=15°.

(3)解:∠OPC=.

证明如下:∵∠POC=∠AOC,∴∠POC=×90°=.∵∠PCE=∠ACE,∴∠PCE=

(180°-45°)=.

∵∠OPC+∠POC=∠PCE,

∴∠OPC=∠PCE-∠POC=.

点睛:本题考查了三角形内角和为180°的性质,考查了角平分线平分角的性质,考查了三角形外角等于不相邻两内角和的性质,本题中求∠PCE和∠POC的大小是解题的关键.

8.如图,已知,在△ABC中,∠B<∠C,AD平分∠BAC,E的线段AD(除去端点A、D)上一动点,EF⊥BC于点F.

(1)若∠B=40°,∠DEF=10°,求∠C的度数.

(2)当E在AD上移动时,∠B、∠C、∠DEF之间存在怎样的等量关系?请写出这个等量关系,并说明理由.

【答案】(1)∠C=60°.

(2)∠C-∠B=2∠DEF.理由见解析

【解析】

试题分析:(1)已知:EF⊥BC,∠DEF=10°可以求得∠EDF的度数,∠EDF又是?ABD的外角,已知∠B的度数,可求得∠BAD的值,AD平分∠BAC,所以∠BAC的值也可求出,从而求出∠C。(2)EF⊥BC,可得到∠EDF=90°-∠DEF,∠EDF又是?ABD的外角,可得到∠BAD=∠EDF-∠B=90°-∠DEF-∠B,然后可将∠ BAC用含∠DEF、∠B的角来表示,即∠BAC =2(90°-∠DEF-∠B),最后利用∠B、∠ BAC、∠C的和为180°求得三角之间的等量关系。

试题解析:(1)∵EF⊥BC,∠DEF=10°,

∴∠EDF=80°.

∵∠B=40°,

∴∠BAD=∠EDF-∠B=80°-40°=40°.

∵AD平分∠BAC,∴∠BAC=80°.

∴∠C=180°-40°-80°=60°.

(2)∠C-∠B=2∠DEF.理由如下:

∵EF⊥BC,∴∠EDF=90°-∠DEF.

∵∠EDF=∠B+∠BAD,

∴∠BAD=90°-∠DEF-∠B.

∵AD 平分∠BAC,

∴∠BAC=2∠BAD=180°-2∠DEF-2∠B. ∴∠B+180°-2∠DEF-2∠B+∠C=180°. ∴∠C -∠B =2∠DEF.

【点睛】本题主要考查考生对三角形外角和性质得理解及灵活运用,以及对三角形内角和,角平分线的定义的理解。此为易考点及重点。考查考生等量代换思想的形成及掌握,在解题过程中涉及到角与角之间的转换。此为难点。

9.如图,在△ABC 中,AD ⊥BC ,AE 平分∠BAC .

(1)若∠B =72°,∠C =30°,①求∠BAE 的度数;②求∠DAE 的度数;

(2)探究:如果只知道∠B =∠C +42°,也能求出∠DAE 的度数吗?若能,请你写出求解过程;若不能,请说明理由.

【答案】(1)①39°;②21°;(2)21°. 【解析】 【分析】

()1①先根据三角形内角和定理计算出BAC 78∠=,然后根据角平分线定义得到

1BAE BAC 392

∠∠==;

②根据垂直定义得到ADB 90∠=,则利用互余可计算出BAD 90B 18∠∠=-=,

然后利用DAE BAE BAD ∠∠∠=-进行计算即可;

()2由B C BAC 180

∠∠∠++=,B C 42∠∠=+可消去C ∠得到

BAC 2222B ∠∠=-,则根据角平分线定义得到BAE 111B ∠∠=-,接着在ABD

中利用互余得BAD 90B ∠∠=-,然后利用DAE BAE BAD ∠∠∠=-进行计算即可得到DAE 21∠=. 【详解】 解:()1B C BAC 180∠∠∠++=①

BAC 180723078∠∴=--=,

AE 平分BAC ∠,

1

BAE BAC 392

∠∠∴==;

AD BC ⊥②, ADB 90∠∴=,

BAD 90B 18∠∠∴=-=,

DAE BAE BAD 391821∠∠∠∴=-=-=;

()2能.

B C BAC 180∠∠∠++=,B C 42∠∠=+,

C B 42∠∠∴=-, 2B BAC 222∠∠∴+=, BAC 2222B ∠∠∴=-,

AE 平分BAC ∠,

BAE 111B ∠∠∴=-,

在ABD 中,BAD 90B ∠∠=-,

()()

DAE BAE BAD 111B 90B 21∠∠∠∠∠∴=-=---=.

【点睛】

本题考查三角形内角和定理:三角形内角和是180.掌握角平分线和高的定义,熟练进行角度的运算.

10.动手操作,探究:

探究一:三角形的一个内角与另两个内角的平分线所夹的钝角之间有何种关系. 已知:如图(1),在△ADC 中,DP 、CP 分别平分∠ADC 和∠ACD ,试探究∠P 与∠A 的数量关系.并说明理由.

探究二:若将△ADC 改为任意四边形ABCD 呢?

已知:如图(2),在四边形ABCD 中,DP 、CP 分别平分∠ADC 和∠BCD ,请你利用上述结论探究∠P 与∠A +∠B 的数量关系,并说明理由.

探究三:若将上题中的四边形ABCD 改为六边形ABCDEF 如图(3)所示,请你直接写出∠P 与∠A +∠B +∠E +∠F 的数量关系.

【答案】探究一: 90°+12∠A ;探究二:1

2

(∠A +∠B );探究三:∠P =

1

2(∠A +∠B +∠E +∠F )﹣180°. 【解析】 试题分析:

探究一:根据角平分线的定义可得∠PDC=1

2

∠ADC,∠PCD=

1

2

∠ACD,然后根据三角

形内角和定理列式整理即可得解.

探究二:根据四边形的内角和定理表示出∠ADC+∠BCD,然后同理探究一解答即可.探究三:根据六边形的内角和公式表示出∠EDC+∠BCD,然后同理探究一解答即可.试题解析:

探究一:∵DP、CP分别平分∠AD C和∠ACD,

∴∠PDC=1

2

∠ADC,∠PCD=

1

2

∠ACD,

∴∠DPC=180°-∠PDC-∠PCD,

=180°-1

2

∠ADC-

1

2

∠ACD,

= 180°-1

2

(∠ADC+∠ACD),

=180°-1

2

(180°-∠A),

=90°+1

2

∠A;

探究二:∵DP、CP分别平分∠ADC和∠BCD,

∴∠PDC=1

2

∠ADC,∠PCD=

1

2

∠BCD,

∴∠DPC=180°-∠PDC-∠PCD,

=180°-1

2

∠ADC-

1

2

∠BCD,

=180°-1

2

(∠ADC+∠BCD),

=180°-1

2

(360°-∠A-∠B),

=1

2

(∠A+∠B);

探究三:六边形ABCDEF的内角和为:(6-2)×180°=720°,∵DP、CP分别平分∠EDC和∠BCD,

∴∠PDC=1

2

∠EDC,∠PCD=

1

2

∠BCD,

∴∠P=180°-∠PDC-∠PCD,

=180°-1

2

∠EDC-

1

2

∠BCD,

=180°-1

2

(∠EDC+∠BCD),

=180°-1

2

(720°-∠A-∠B-∠E-∠F),

=1

2

(∠A+∠B+∠E+∠F)-180°,

即∠P=1

2

(∠A+∠B+∠E+∠F)-180°.

点睛:本题考查了三角形的外角性质,三角形的内角和定理,多边形的内角和公式,在此类题目中根据同一个解答思路求解是解题的关键.

2020-2021上海民办上宝中学小学数学小升初模拟试卷(含答案)

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上海市上宝中学数学圆 几何综合章末训练(Word版 含解析)

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2017-2018上海上宝中学九上英语期中考试(附答案)

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2017-2018学年上宝中学九上英语12月月考 上海市 英语试卷

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