2026年学习检测七年级数学下册华师大版河南专版第70页答案
8. (聊城期末)如图,在$△ ABC$中,$BD$是边$AC$上的高,$CE$是$∠ ACB$的平分线,$BD$、$CE$交于点$F$. 若$∠ AEC = 80°$,$∠ BFC = 128°$,则$∠ ABC$的度数是【 】


A.$28°$
B.$38°$
C.$42°$
D.$62°$

答案

8. C
9. 有下列条件:①$∠ A + ∠ B = ∠ C$;②$∠ A:∠ B:∠ C = 5:3:2$;③$∠ A = 90° - ∠ B$;④$∠ A = 2∠ B = 3∠ C$;⑤$∠ A = \frac{1}{2}∠ B = \frac{1}{3}∠ C$. 其中,能确定$△ ABC$是直角三角形的有【 】

A.$2$个
B.$3$个
C.$4$个
D.$5$个

答案

9. C
10. 在$△ ABC$中,$∠ B = 40°$,$∠ C = 60°$,$∠ B$和$∠ C$的平分线交于点$O$,则$∠ BOC =$
$130°$
.

答案

10. $130°$
11. 如图,$AD$是$△ ABC$的角平分线,$AE ⊥ BC$于点$E$. 若$∠ B = 40°$,$∠ C = 70°$,则$∠ DAE$的度数为
15
$°$.

答案

11. 15
12. 如图,在$△ ABC$中,$AD$平分$∠ BAC$,点$E$在$AD$上,$EF ⊥ AD$交$BC$的延长线于点$F$. 若$∠ B = 40°$,$∠ ACB = 70°$,求$∠ F$的度数.

答案

12. $∠ BAC = 180°-∠ ACB-∠ B = 180°-40°-70°=70°$,$\therefore∠ CAD=\frac{1}{2}∠ BAC = 35°$。$\therefore∠ ADC = 180°-∠ CAD-∠ ACB = 75°$。$\therefore∠ F = 90°-∠ ADC = 15°$
13. (洛阳期中)如图,在$△ ABC$中,已知$AD$是$△ ABC$的角平分线,$DE$是$△ ADC$的高. 若$∠ B = 60°$,$∠ C = 40°$,求$∠ ADB$和$∠ ADE$的度数.

答案

13. $\because$在$△ ABC$中,$∠ B = 60°$,$∠ C = 40°$,$\therefore∠ BAC = 80°$。$\because AD$是$△ ABC$角平分线,$\therefore∠ BAD=∠ DAC=\frac{1}{2}∠ BAC = 40°$。$\therefore∠ ADB = 80°$。$\because DE$是$△ ADC$的高,$\therefore∠ DEA = 90°$,$\therefore∠ ADE = 50°$