如图,BD平分∠ABC,BE把∠ABC分成2∶5的两部分,∠DBE=21°,求∠ABC的度数.A ED BC 答案 解:设∠ABE=2°,则∠CBE=5°,∠ABC=7°.(1分)又因为BD为∠ABC的平分线,所以∠ABD=2∠ABC=72°,(2分)∠DBE=∠ABD-∠ABE=72°-2°=32°=21°.(3分)所以=14,所以∠ABC=7°=98°.(6分) 相关...
【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$,$CD$平分$\angle BCA$,若$\angle D=3\angle A$,则$\angle A=\left( \right)$ A.$32^{\circ}$ B.$ 36^{\circ}$ C.$ 40^{\circ}$ D.$ 44^{\circ}$ 相关知识点: 试题来源: ...
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}$ ...
【答案】见解析.【解析】$\because $$BD$平分$\angle ABC$,$CE$平分$\angle ACB$,$\therefore $$\angle ABD=\angle DBC$,$\angle ACE=\angle ECB$,$\because $$\angle ABC=\angle ACB$,$\therefore $$\angle DBC=\angle ECB$,$\because $$\angle DBF=\angle F$,$\therefore $$\angle DBC...
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^{\...
又∵ BD是∠ ABC的平分线,∴ DE=DF,又∵ ∠ BAD+∠ CAD=180°,∠ BAD+∠ EAD=180°,∴∠ CAD=∠ EAD,∴ AD为∠ EAC的平分线,∴ DE=DG,∴ DG=DF.∴ CD为∠ ACF的平分线,∵∠ DCB=117°,∴∠ DCF=63°,∴∠ ACF=126°,∴∠ BAC=∠ ACF-∠ ABC=126°-50°=76°,∴∠ CAE=104°,...
1如图,四边形$ABCD$,$BP$、$CP$分别平分$\angle ABC$、$\angle BCD$,写出$\angle A$、$\angle D$、$\angle P$之间的数量关系为___. 2如图,四边形$ABCD$,$BP$、$CP$分别平分$\angle ABC$、$\angle BCD$,写出$\angle A$、$\angle D$、$\angle P$之间的数量关系为___. 3如图,四边形$...
∵ BD平分∠ ABC,∠ C=90°,∴ CD=DE,在Rt△ BCD与Rt△ BED中\((array)l(CD=DE)(BD=BD)(array).,∴ Rt△ BCD≌Rt△ BED(HL),∴ BE=BC=4,∴ AE=1,∵ AD^2=DE^2+AE^2,∴ AD^2=(3-AD)^2+1^2,∴ AD=5/3,故答案为:5/3....
如图,$\triangle ABC$中,$AB=AC$,$\angle A=36^{\circ}$,$BD$平分$\angle ABC$,且$DE$∥$B C.$那么与$