《2025年基础训练大象出版社八年级数学上册人教版》

18. (9分)如图,在$\triangle ABC$中,$∠ABC = 30^{\circ}$,$∠ACB = 50^{\circ}$,DE,FG分别为AB,AC的垂直平分线,E,G分别为垂足.
(1)求$∠DAF$的度数;
(2)若$\triangle DAF$的周长为20,求BC的长.

答案：
(1)
∵ ∠ABC + ∠ACB + ∠BAC = 180°，∠ABC = 30°，∠ACB = 50°，
∴ ∠BAC = 180° - 30° - 50° = 100°.
∵ DE是线段AB的垂直平分线，
∴ DA = DB.
∴ ∠DAB = ∠ABC = 30°.
同理可得，FA = FC，∠FAC = ∠ACB = 50°.
∴ ∠DAF = ∠BAC - ∠DAB - ∠FAC = 100° - 30° - 50° = 20°.
(2)
∵ △DAF的周长为20，
∴ DA + DF + FA = 20.
由
(1)可知，DA = DB，FA = FC，
∴ BC = DB + DF + FC = DA + DF + FA = 20.

19. (9分)如图,在$\triangle ABC$中,$AB = AC$,点D在边BC上,连接AD.分别过点A,C作$AF ⊥ BC$,$CE ⊥ AD$,垂足分别是F,E.已知$∠ACE = \frac{1}{2}∠BAC$.
求证:(1)$∠ACE = ∠CAF$;
(2)$AD = CD$.

答案：
(1)
∵ AB = AC，AF⊥BC，
∴ ∠CAF = ∠BAF = $\frac{1}{2}$∠BAC.
∵ ∠ACE = $\frac{1}{2}$∠BAC，
∴ ∠ACE = ∠CAF;
(2)
∵ CE⊥AD，AF⊥CB，
∴ ∠CEA = ∠AFC = 90°.
在△ACE和△CAF中，
$\begin{cases}∠CEA = ∠AFC, \\∠ACE = ∠CAF, \\AC = CA,\end{cases}$
∴ △ACE≌△CAF(AAS).
∴ ∠DAC = ∠ACD.
∴ AD = CD.