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八年级数学解答题一般
题目
如图,AAEEBB三点在一条直线上,ACBCAC\bot BC,ADDBAD\bot DB,BC=BDBC=BD.求证:CE=DECE=DE.
知识点:等腰三角形的性质、勾股定理、直角三角形的性质章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

证明:ACBC\because AC\bot BCADDBAD\bot DB
ACB=ADB=90\therefore \angle ACB=\angle ADB=90^{\circ}
RtABCRt\triangle ABCRtABDRt\triangle ABD中,
{AB=ABBC=BD\left\{\begin{array}{l}{AB=AB}\\{BC=BD}\end{array}\right.
RtABC\therefore Rt\triangle ABCRtABD(HL)Rt\triangle ABD \left(HL\right)
ABC=ABD\therefore \angle ABC=\angle ABD
EBC\triangle EBCEBD\triangle EBD中,
{BC=BDEBC=EBDBE=BE\left\{\begin{array}{l}{BC=BD}\\{∠EBC=∠EBD}\\{BE=BE}\end{array}\right.
EBC\therefore \triangle EBCEBD(SAS)\triangle EBD \left(SAS\right)
CE=DE\therefore CE=DE.

解析

证明:ACBC\because AC\bot BCADDBAD\bot DB
ACB=ADB=90\therefore \angle ACB=\angle ADB=90^{\circ}
RtABCRt\triangle ABCRtABDRt\triangle ABD中,
{AB=ABBC=BD\left\{\begin{array}{l}{AB=AB}\\{BC=BD}\end{array}\right.
RtABC\therefore Rt\triangle ABCRtABD(HL)Rt\triangle ABD \left(HL\right)
ABC=ABD\therefore \angle ABC=\angle ABD
EBC\triangle EBCEBD\triangle EBD中,
{BC=BDEBC=EBDBE=BE\left\{\begin{array}{l}{BC=BD}\\{∠EBC=∠EBD}\\{BE=BE}\end{array}\right.
EBC\therefore \triangle EBCEBD(SAS)\triangle EBD \left(SAS\right)
CE=DE\therefore CE=DE.

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