《2025年新课标教材同步导练八年级数学下册北师大版C版》

17. (2021·绍兴）如图，在△ABC中，∠A = 40°，点D、E分别在边AB、AC上，BD = BC = CE，连接CD、BE.
(1）若∠ABC = 80°，求∠BDC，∠ABE的度数；
(2）写出∠BEC与∠BDC之间的关系，并说明理由.

答案：
(1)$\because \angle ABC = 80^{\circ}$，$BD = BC$，
$\therefore \angle BDC = \angle BCD = \frac{1}{2}(180^{\circ} - 80^{\circ}) = 50^{\circ}$，
$\because \angle A + \angle ABC + \angle ACB = 180^{\circ}$，$\angle A = 40^{\circ}$，
$\therefore \angle ACB = 180^{\circ} - 40^{\circ} - 80^{\circ} = 60^{\circ}$，
$\because CE = BC$，$\therefore \triangle BCE$是等边三角形，
$\therefore \angle EBC = 60^{\circ}$，
$\therefore \angle ABE = \angle ABC - \angle EBC = 20^{\circ}$。
(2)$\angle BEC$与$\angle BDC$之间的关系：$\angle BEC + \angle BDC = 110^{\circ}$，
理由：设$\angle BEC = \alpha$，$\angle BDC = \beta$，
在$\triangle ABE$中，$\alpha = \angle A + \angle ABE = 40^{\circ} + \angle ABE$，
$\because CE = BC$，$\therefore \angle CBE = \angle BEC = \alpha$，
$\therefore \angle ABC = \angle ABE + \angle CBE = \angle A + 2\angle ABE = 40^{\circ} + 2\angle ABE$，
在$\triangle BDC$中，$BD = BC$，$\therefore \angle BDC = \angle BCD = \beta$，
$\therefore \angle BDC + \angle BCD + \angle DBC = 2\beta + 40^{\circ} + 2\angle ABE = 180^{\circ}$，
$\therefore \beta = 70^{\circ} - \angle ABE$，
$\therefore \alpha + \beta = 40^{\circ} + \angle ABE + 70^{\circ} - \angle ABE = 110^{\circ}$，
$\therefore \angle BEC + \angle BDC = 110^{\circ}$。

18. 如图，已知点D为等腰直角△ABC内一点，∠ACB = 90°，∠CAD = ∠CBD = 15°，E为AD延长线上的一点，且CE = CA.
(1）求证：DE平分∠BDC；
(2）若点M在DE上，且DC = DM，求证：ME = BD.

答案：
(1)在等腰直角$\triangle ABC$中，$\angle ACB = 90^{\circ}$，
$\because \angle CAD = \angle CBD = 15^{\circ}$，
$\therefore \angle BAD = \angle ABD = 45^{\circ} - 15^{\circ} = 30^{\circ}$，
$\therefore BD = AD$，$\therefore \triangle BDC\cong\triangle ADC$，
$\therefore \angle DCA = \angle DCB = 45^{\circ}$。
又$\angle BDM = \angle ABD + \angle BAD = 30^{\circ} + 30^{\circ} = 60^{\circ}$，
$\angle EDC = \angle DAC + \angle DCA = 15^{\circ} + 45^{\circ} = 60^{\circ}$，
$\therefore \angle BDM = \angle EDC$，$\therefore DE$平分$\angle BDC$。
(2)连接$MC$。$\because DC = DM$，且$\angle MDC = 60^{\circ}$，
$\therefore \triangle MDC$是等边三角形，即$CM = CD$。
又$\angle EMC = 180^{\circ} - \angle DMC = 180^{\circ} - 60^{\circ} = 120^{\circ}$，$\angle ADC = 180^{\circ} - \angle MDC = 180^{\circ} - 60^{\circ} = 120^{\circ}$，$\therefore \angle EMC = \angle ADC$。
又$\because CE = CA$，$\therefore \angle DAC = \angle CEM = 15^{\circ}$，
$\therefore \triangle ADC\cong\triangle EMC$，$\therefore ME = AD = BD$。