《2025年拔尖特训八年级数学上册人教版》

10. 如图，在$\triangle ABC$中，点$D$，$E$分别在边$AB$，$BC$上，连接$AE$，$DE$. 若$\triangle ADE≌\triangle BDE$，$AC:AB:BC=2:3:4$，且$\triangle ABC$的周长比$\triangle AEC$的周长大$6$，求$\triangle AEC$的周长.

答案： $\because \triangle ADE \cong \triangle BDE$，
$\therefore AE = BE$。
$\therefore C_{\triangle AEC} = AE + EC + AC = BE + EC + AC = BC + AC$。
$\because AC:AB:BC = 2:3:4$，
$\therefore$ 设 $AC = 2x$，则 $AB = 3x$，$BC = 4x$。
$\because \triangle ABC$ 的周长比 $\triangle AEC$ 的周长大 6，
$\therefore C_{\triangle ABC} - C_{\triangle AEC} = 6$。
$\therefore (AB + BC + AC) - (BC + AC) = 6$。
$\therefore AB = 3x = 6$，解得 $x = 2$。
$\therefore AC = 2x = 4$，$BC = 4x = 8$。
$\therefore C_{\triangle AEC} = BC + AC = 8 + 4 = 12$。

11. 如图，$\triangle BKC≌\triangle BKE≌\triangle DKC$，$BE$与$KD$交于点$G$，$KE$与$CD$交于点$P$，$BE$与$CD$交于点$A$，$\angle BKC=135^{\circ}$，$\angle E=22^{\circ}$，求$\angle KPD$的度数.

答案： $\because \triangle BKC \cong \triangle BKE \cong \triangle DKC$，$\angle BKC = 135^{\circ}$，$\angle E = 22^{\circ}$，
$\therefore \angle DCK = \angle E = 22^{\circ}$，$\angle BKE = \angle DKC = \angle BKC = 135^{\circ}$。
$\therefore \angle DKP = \angle BKC + \angle DKC + \angle BKE - 360^{\circ} = 45^{\circ}$。
$\therefore \angle EKC = \angle DKC - \angle DKE = 135^{\circ} - 45^{\circ} = 90^{\circ}$。
$\therefore \angle KPD = \angle PCK + \angle PKC = 22^{\circ} + 90^{\circ} = 112^{\circ}$。

12. 如图，在锐角三角形$ABC$中，$D$，$E$分别是边$AB$，$AC$上的点，$\triangle ADC≌\triangle ADC'$，$\triangle AEB≌\triangle AEB'$，且$C'D// EB'// BC$，$BE$，$CD$交于点$F$. 若$\angle BAC=\alpha$，$\angle BFC=\beta$，则 （ ）

A. $2\alpha+\beta=180^{\circ}$
B. $2\beta-\alpha=180^{\circ}$
C. $\alpha+\beta=150^{\circ}$
D. $\beta-\alpha=60^{\circ}$

答案：
A 解析：如图，延长 $C'D$ 交 $AC$ 于点 $M$。$\because \triangle ADC \cong \triangle ADC'$，$\triangle AEB \cong \triangle AEB'$，$\therefore \angle ACD = \angle C'$，$\angle ABE = \angle B'$，$\angle CAD = \angle C'AD = \angle B'AE = \alpha$。$\therefore \angle C'MC = \angle C' + \angle C'AM = \angle C' + 2\alpha$。
$\because C'D // EB'$，$\therefore \angle AEB' = \angle C'MC$。
$\because \angle AEB' = 180^{\circ} - \angle B' - \angle B'AE = 180^{\circ} - \angle B' - \alpha$，$\therefore \angle C' + 2\alpha = 180^{\circ} - \angle B' - \alpha$。$\therefore \angle C' + \angle B' = 180^{\circ} - 3\alpha$。$\because \angle BFC = \beta = \angle BDF + \angle DBF$，$\angle BDF = \angle DAC + \angle ACD$，
$\therefore \angle BFC = \beta = \angle DAC + \angle ACD + \angle B' = \alpha + \angle ACD + \angle B' = \alpha + \angle C' + \angle B' = \alpha + 180^{\circ} - 3\alpha = 180^{\circ} - 2\alpha$，即 $2\alpha + \beta = 180^{\circ}$。

13. 如图，$\triangle ABE≌\triangle ADC≌\triangle ABC$，若$\angle 1:\angle 2:\angle 3=13:3:2$，$CD$与$BE$交于点$O$，求$\angle EOC$的度数.

答案： 设 $AE$ 与 $DC$ 交于点 $P$。
$\because \angle 1:\angle 2:\angle 3 = 13:3:2$，
$\therefore$ 易得 $\angle 1 = 130^{\circ}$，$\angle 2 = 30^{\circ}$，$\angle 3 = 20^{\circ}$。
$\because \triangle ABE \cong \triangle ADC \cong \triangle ABC$，
$\therefore \angle DCA = \angle E = \angle 3 = 20^{\circ}$，$\angle EAB = \angle 1 = 130^{\circ}$。
$\therefore \angle PAC = 360^{\circ} - 2\angle 1 = 100^{\circ}$。
$\therefore \angle EPD = \angle APC = 180^{\circ} - \angle PAC - \angle DCA = 60^{\circ}$。
$\therefore \angle EOC = 180^{\circ} - \angle EPD - \angle E = 180^{\circ} - 60^{\circ} - 20^{\circ} = 100^{\circ}$。