8 如图,四边形 ABCD 内接于$\odot O$,点 I 是$△ ABC$的内心,$∠ AIC=124°$,点 E 在 AD 的延长线上,则$∠ CDE$的度数为 ( )
A. $56°$
B. $62°$
C. $68°$
D. $78°$



A. $56°$
B. $62°$
C. $68°$
D. $78°$
答案
C
9 如图,点 I 为$△ ABC$的内心,$AB=4$,$AC=3$,$BC=2$,将$∠ ACB$平移使其顶点与点 I 重合,则图中阴影部分的周长为______.
答案
4
10 如图,在扇形$CAB$中,$CD\bot AB$,垂足为$D$,$\odot E$是$△ ACD$的内切圆,连接$AE$,$BE$,则$∠ AEB$的度数为______.
答案
135°
11 如图,BC为$△ ABC$的外接圆$\odot O$的直径,点M为$△ ABC$的内心,连接AM并延长交$\odot O$于点D,连接CD.
(1)求$∠ BCD$的大小;
(2)若$CD=4$,求$DM$的值.

(1)求$∠ BCD$的大小;
(2)若$CD=4$,求$DM$的值.
答案
解:$(1)$∵$BC $为$∆ABC $的外接圆$⊙O $的直径,∴$∠BAC = 90°$∵点$ M $为$∆ABC $的内心,∴$∠BAD = \frac 12∠BAC = 45°$∴$∠BCD = ∠BAD = 45°$$ (2) $连接$ CM$∵点$ M $为$∆ABC $的内心∴$∠BAD = ∠CAD,$$∠ACM = ∠BCM$∵$∠BAD = ∠BCD,$∴$∠DAC = ∠BCD$∵$∠DMC = ∠DAC + ∠ACM,$$∠DCM = ∠BCD + ∠BCM$∴$∠DMC = ∠DCM$∴$DM = CD = 4$ ;
12 如图,AB为$\odot O$的直径,$△ ABC$内接于$\odot O$,$BC>AC$,$\overset{\frown}{AD}=\overset{\frown}{BD}$,$BD=PD$,延长$CP$交$\odot O$于点$D$,连接$BP$.
(1) 求证:点$P$是$△ ABC$的内心;
(2) 已知$\odot O$的直径是$5\sqrt{2}$,$CD=7$,求$BC$的长.

(1) 求证:点$P$是$△ ABC$的内心;
(2) 已知$\odot O$的直径是$5\sqrt{2}$,$CD=7$,求$BC$的长.
答案
$ (1) $证明:∵$AB $为$⊙O $的直径,∴$∠ACB = 90°$∵$\widehat {AD}=\widehat {BD}$∴$∠ACD = ∠BCD,$∴$CD $平分$∠ACB$∵$P D = BD,$∴$∠BP D = ∠P BD$∵$∠BP D = ∠BCP + ∠CBP,$$∠DBP = ∠ABD + ∠ABP,$$∠ABD = ∠DCB$∴$∠CBP = ∠ABP,$∴$BP $平分$∠ABC$∴点$ P $是$∆ABC $的内心$ (2) $解:如图,连接$ AD,$过点$ B $作$ BH⊥CD $于点$ H$$ $易知$∆ABD $是等腰直角三角形∴$BD = \frac {\sqrt 2}2\ \mathrm {A}B = \frac {\sqrt 2}2×5\sqrt 2 = 5$∵$∠BCD = 45°,$$BH⊥CD$∴$∠BCH = ∠CBH = 45°$∴$BH = CH,$∴$BC = \sqrt 2BH$∵$BD^2 = DH^2 + BH^2$∴$25 = (7 - BH)^2 + BH^2,$解得$ BH = 3 $或$ BH = 4$ 又∵$BC>AC,$∴$BH = 4,$∴$BC = 4\sqrt 2$ ;
登录