一道数学题.不是很难的.(1)如图,在△ABC中,∠ACB=100°,AC=AE,BC=BD,则∠DEC的度数为( )A.20° B.25° C.30° D.40°

来源:学生作业帮助网 编辑:作业帮 时间:2024/07/11 23:32:48
一道数学题.不是很难的.(1)如图,在△ABC中,∠ACB=100°,AC=AE,BC=BD,则∠DEC的度数为( )A.20° B.25° C.30° D.40°
xRn@+RW+Q\)~|vcXJHhi$]R6 d6 >%ɪtU`;Ιs\O)\K'o{}N$~9Ji{3gMEOIgHϣ˺=z<,!O lѺ&Maϴ  .'(*3<)%(#ddD(%g6ONסr뒕u ߳H+w;h kJbYaZFcW7Gʊ`Z !6 75uUI,ǡ"s6]Uq>\OyaIdUVNQe,Kqőx$8XV}U|G *y/0 J5,I[e22c03覽֤O;$nw$=i'NNg+ ,rsX

一道数学题.不是很难的.(1)如图,在△ABC中,∠ACB=100°,AC=AE,BC=BD,则∠DEC的度数为( )A.20° B.25° C.30° D.40°
一道数学题.不是很难的.
(1)如图,在△ABC中,∠ACB=100°,AC=AE,BC=BD,则∠DEC的度数为( )
A.20° B.25° C.30° D.40°

一道数学题.不是很难的.(1)如图,在△ABC中,∠ACB=100°,AC=AE,BC=BD,则∠DEC的度数为( )A.20° B.25° C.30° D.40°
∵AC=AE,BC=BD
∴∠ACE=∠AEC,∠BCD=∠BDC(等腰直角三角形的两底角相等且都等于45°)
∵∠ACB=100°
∴∠ACE+∠BCD=∠AEC+∠BDC=100°+∠DCE ①
∵在△DCE中,由三角形内角和定理有
∴∠AEC+∠BDC+∠DCE=180° ②
将①代入②,得
∴100°+∠DCE+∠DCE=180°
解得∠DCE=40°

D

D

D