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八年级数学解答题一般
题目
如图所示,在等腰ABC\triangle ABC中,AB=ACAB=AC,点DD,EE,FFABC\triangle ABC的边上,满足BE=CFBE=CF,BD=CEBD=CE.
(1)(1)求证:DE=EFDE=EF
(2)(2)A=80\angle A=80^{\circ}时,求EDF\angle EDF的大小.
知识点:角的运算、三角形内角和定理、全等三角形的性质、全等三角形的判定、等腰三角形的性质、勾股定理、旋转的性质章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

(1)(1)证明:AB=AC\because AB=AC
B=C\therefore \angle B=\angle C
BDE\triangle BDECEF\triangle CEF中,
{BE=CFB=CBD=CE\left\{\begin{array}{l}{BE=CF}\\{∠B=∠C}\\{BD=CE}\end{array}\right.
BDE\therefore \triangle BDECEF(SAS)\triangle CEF\left(SAS\right)
DE=EF\therefore DE=EF.
(2)(2)A=80\because \angle A=80^{\circ}
B=C=12×(18080)=50\therefore \angle B=\angle C=\frac{1}{2}\times \left(180^{\circ}-80^{\circ}\right)=50^{\circ}
BDE\because \triangle BDECEF\triangle CEF
BDE=CEF\therefore \angle BDE=\angle CEF
DEF=180BEDCEF=180BEDBDE=B=50\therefore \angle DEF=180^{\circ}-\angle BED-\angle CEF=180^{\circ}-\angle BED-\angle BDE=\angle B=50^{\circ}
DE=DF\because DE=DF
EDF=EFD=12×(18050)=65\therefore \angle EDF=\angle EFD=\frac{1}{2}\times \left(180^{\circ}-50^{\circ}\right)=65^{\circ}
EDF\therefore \angle EDF的度数是6565^{\circ}.

解析

(1)(1)证明:AB=AC\because AB=AC
B=C\therefore \angle B=\angle C
BDE\triangle BDECEF\triangle CEF中,
{BE=CFB=CBD=CE\left\{\begin{array}{l}{BE=CF}\\{∠B=∠C}\\{BD=CE}\end{array}\right.
BDE\therefore \triangle BDECEF(SAS)\triangle CEF\left(SAS\right)
DE=EF\therefore DE=EF.
(2)(2)A=80\because \angle A=80^{\circ}
B=C=12×(18080)=50\therefore \angle B=\angle C=\frac{1}{2}\times \left(180^{\circ}-80^{\circ}\right)=50^{\circ}
BDE\because \triangle BDECEF\triangle CEF
BDE=CEF\therefore \angle BDE=\angle CEF
DEF=180BEDCEF=180BEDBDE=B=50\therefore \angle DEF=180^{\circ}-\angle BED-\angle CEF=180^{\circ}-\angle BED-\angle BDE=\angle B=50^{\circ}
DE=DF\because DE=DF
EDF=EFD=12×(18050)=65\therefore \angle EDF=\angle EFD=\frac{1}{2}\times \left(180^{\circ}-50^{\circ}\right)=65^{\circ}
EDF\therefore \angle EDF的度数是6565^{\circ}.

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