第11页 - 通城学典课时作业本八年级数学苏科版江苏专版

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 $\triangle AOD\cong\triangle AOE，$$\triangle DOC\cong\triangle EOB，$$\triangle AOC\cong\triangle AOB，$$\triangle ACE\cong\triangle ABD$

证明：∵$AB// DE，$∴$∠CAB = ∠E$

∵$∠E = 40°，$∴$∠CAB = 40°$

∵$∠DAB = 70°，$∴$∠DAE = ∠DAB - ∠CAB = 30°$

∵$∠B = 30°，$∴$∠DAE = ∠B$

在$\triangle ADE$和$\triangle BCA$中

$\begin {cases}∠DAE = ∠B \\AE = BA \\∠E = ∠CAB\end {cases}$

∴$\triangle ADE\cong \triangle BCA(AS A)，$∴$AD = BC$

解：$BC// DF，$$BC = DF，$理由如下

∵$∠ABC + ∠CBD = 180°，$$∠EDF + ∠F DB = 180°，$$∠ABC = ∠EDF$

∴$∠CBD = ∠F DB，$∴$BC//DF$

∵$AD = BE，$∴$AD - BD = BE - BD，$即$AB = ED$

∵$AC// EF，$∴$∠A = ∠E$

在$\triangle ABC$和$\triangle EDF {中}$

$\begin {cases}∠A = ∠E \\AB = ED \\∠ABC = ∠EDF\end {cases}$

∴$\triangle ABC\cong \triangle EDF(AS A)$

∴$BC = DF$

解：∵$BE\perp CE，$$AD\perp CE，$∴$∠E = ∠ADC = 90°$

∴$∠EBC + ∠BCE = 90°，$$∠CAD + ∠DCA = 90°$

∵$∠ACB = ∠BCE + ∠DCA = 90°$

∴$∠EBC = ∠DCA，$$∠BCE = ∠CAD$

$ $在$\triangle CEB$和$\triangle ADC$中

$ \begin {cases}∠EBC = ∠DCA \\BC = CA \\∠BCE = ∠CAD\end {cases}$

∴$\triangle CEB≌\triangle ADC(AS A)$

∴$BE = CD，$$CE = AD$

∵$BE = 1，$$AD = 3，$∴$CD = 1，$$CE = 3$

∴$DE = CE - CD = 3 - 1 = 2$