【答案】$142$【解析】$\because BD$平分$\angle ABC$,$\therefore \angle ABD=\angle DBC$,设$\angle ABD=x$,$DE$与$BC$交于点$M$,$AE$与$BD$交于点$G$.$\because \angle AGB=\angle DGE$,$\because \angle AGB={180}^{\circ }-\angle A-\angle ABD$,$\angle DGE={180}^{\circ...
如图,在$\triangle ABC$中,$BD$平分$\angle ABC$,$BC$的垂直平分线交$BC$于点$E$,交$BD$于点$F$,连接$CF$.若$\angle A=60^{\circ}$,$\angle ABD=20^{\circ}$,则$\angle ACF$的度数为( )A DF BE CA.$60^{\circ}$ B.$ 50^{\circ}$ C.$ 40^{\circ}$ D.$ 20^{\circ}$ ...
1如图,在$\triangle ABC$中,$BD$平分$\angle ABC$,$DE//BC$,且交$AB$于点$E$,$\angle A=60{}^\circ $,$\angle BDC=86{}^\circ $,则$\angle BDE$的度数为 ___ . 2如图,在$\triangle ABC$中,$BD$平分$\angle ABC$,$DE\parallel BC$,且交$AB$于点$E$,$\angle A=60^{\...
4.如图3,{\angle}A=40^{\circ},{\angle}C=80^{\circ},BD平分{\angle}ABC,DE/\!/CB,交AB于E,则{\angle}BDE=___.相关知识点: 试题来源: 解析 30^o 根据三角形内角和性质,可求得∠ ABC=60°。由于BD平分∠ ABC,所以∠ DBC=30°。又因为DE//CB,所以∠ BDE=∠ DBC=30°。因此,∠ BD...
①$\because BD$平分$\angle ABC$,$BM$是$\angle ABC$的外角平分线,$\therefore \angle MBF=\angle MBH=\dfrac{1}{2}\angle ABH$,$\angle ABD=\angle CBD=\dfrac{1}{2}\angle ABC$,$\therefore \angle MBD=\angle MBF+\angle ABD=\dfrac{1}{2}\left(\angle ABH+\angle ABC\right)=\df...
【答案】$D$【解析】$\because AF\ykparallel CD$,$\therefore \angle ABC=\angle ECB$,$\angle EDB=\angle DBF$,$\angle DEB=\angle EBA$,$\because CB$平分$\angle ACD$,$BD$平分$\angle EBF$,$\therefore \angle ECB=\angle BCA$,$\angle EBD=\angle DBF$,$\because BC\bot BD$,$\the...
如图\(①\),\(\Delta ABC\)的角平分线\(BD\)、\(CE\)相交于点\(P\).E A△A PP B◇B①②③ \((1)\)如果\(\angle A={{80}^{\circ }}\),求\(∠BPC\)的度数; \((2)\)如图\(②\),作\(\Delta ABC\)外角\(\angle MBC,\angle NCB\)的角平分线交于点\(Q\),试探索\...
{\circ}$,$\therefore \angle ABD=\angle BCF$,$\because BD$是$\angle ABC$的角平分线,$\therefore \angle ABD=\angle DBC$,$\therefore \angle DBC=\angle BCF$,$\therefore DE$∥$CF.$$(3)$由(2)知$,DE$∥$CF$,$\therefore \angle ADB=\angle ACF$,$\because CB$是$\angle ACF$...
解析 40 根据角平分线性质,可知 ∠ ABD = 1/2 ∠ A,∠ ACD = 1/2 ∠ ACB。由三角形内角和定理,可得 ∠ BDC = 180° - ∠ CBD - ∠ BCD = 180° - 1/2 (∠ ABC + ∠ ACB) = 90° + 1/2 ∠ A。代入 ∠ BDC = 110°,解得 ∠ A = 40°。
{\circ }$;$\because BP$、$CP$分别是$\angle ABC$与$\angle ACB$的外角平分线,$\therefore \angle CBP=\dfrac {1} {2}\angle CBM$,$\angle BCP=\dfrac {1} {2}\angle BCN$,$\therefore \angle CBP+\angle BCP=\dfrac {1} {2}\angle CBM+\dfrac {1} {2}\angle BCN$$=\dfrac {1} ...