《2025年真题圈八年级数学下册北师大版》

18.(月考·22 - 23山西省实验)(8分)如图，AC，BD是□ABCD的两条对角线，AE⊥BD，CF⊥BD，垂足分别为E，F. 求证：EO = FO.

答案： [证明]
∵四边形$ABCD$是平行四边形，
∴$AO = CO$.
∵$AE \perp BD$，$CF \perp BD$，
∴$\angle AEO = \angle CFO = 90^{\circ}$.
在$\triangle AOE$和$\triangle COF$中，$\angle AEO = \angle CFO$，$\angle AOE = \angle COF$，$AO = CO$，
∴$\triangle AOE \cong \triangle COF(AAS)$，
∴$EO = FO$.

19.(联考·21 - 22运城三校)(8分)如图，在五边形ABCDE中，AE//BC，EF平分∠AED，CF平分∠BCD，若∠EDC = 90°，求∠EFC的度数.

答案： [解]
∵$EF$平分$\angle AED$，$CF$平分$\angle BCD$，
∴$\angle AEF = \angle DEF = \frac{1}{2}\angle AED$，$\angle BCF = \angle DCF = \frac{1}{2}\angle BCD$.
∵$AE // BC$，
∴$\angle A + \angle B = 180^{\circ}$.
∵五边形的内角和为$(5 - 2) \times 180^{\circ} = 540^{\circ}$，$\angle D = 90^{\circ}$，
∴$\angle AED + \angle BCD = 540^{\circ} - (\angle A + \angle B + \angle D) = 540^{\circ} - (180^{\circ} + 90^{\circ}) = 270^{\circ}$，
∴$\angle DEF + \angle DCF = \frac{1}{2}(\angle AED + \angle BCD) = \frac{1}{2} \times 270^{\circ} = 135^{\circ}$.
∵四边形$EFCD$的内角和为$360^{\circ}$，
∴$\angle EFC = 360^{\circ} - (\angle D + \angle DEF + \angle DCF) = 360^{\circ} - (90^{\circ} + 135^{\circ}) = 135^{\circ}$.