3. 如图 13 - 19,在$Rt\triangle ABC$中,$∠ACB = 90^{\circ}$,$∠A = 40^{\circ}$,$\triangle ABC的外角∠CBD的平分线BE交AC的延长线于点E$。
(1)求$∠CBE$的度数;
(2)过点$D作DF// BE$,交$AC的延长线于点F$,求$∠F$的度数。
(1)求$∠CBE$的度数;
65°
(2)过点$D作DF// BE$,交$AC的延长线于点F$,求$∠F$的度数。
25°
答案
3.(1)∵ 在$Rt△ABC$中,$∠ACB = 90^{\circ}$,$∠A = 40^{\circ}$,
∴$∠ABC = 90^{\circ} - ∠A = 50^{\circ}$.
∴$∠CBD = 130^{\circ}$.
∵ BE 是$∠CBD$的平分线,
∴$∠CBE = \frac{1}{2}∠CBD = 65^{\circ}$.
(2)∵$∠ACB = 90^{\circ}$,$∠CBE = 65^{\circ}$,
∴$∠CEB = 90^{\circ} - 65^{\circ} = 25^{\circ}$.
∵$DF // BE$,∴$∠F = ∠CEB = 25^{\circ}$.
∴$∠ABC = 90^{\circ} - ∠A = 50^{\circ}$.
∴$∠CBD = 130^{\circ}$.
∵ BE 是$∠CBD$的平分线,
∴$∠CBE = \frac{1}{2}∠CBD = 65^{\circ}$.
(2)∵$∠ACB = 90^{\circ}$,$∠CBE = 65^{\circ}$,
∴$∠CEB = 90^{\circ} - 65^{\circ} = 25^{\circ}$.
∵$DF // BE$,∴$∠F = ∠CEB = 25^{\circ}$.
4. 如图 13 - 20,在$\triangle ABC$中,$BE$是角平分线,点$D在边AB$上(不与点$A$,$B$重合),$CD与BE交于点O$。

(1)若$CD$是中线,$BC = 3$,$AC = 2$,则$\triangle BCD与\triangle ACD$的周长差为______
(2)若$CD$是高,$∠ABC = 62^{\circ}$,求$∠BOC$的度数;
(3)若$CD$是角平分线,$∠A = 78^{\circ}$,求$∠BOC$的度数。
(1)若$CD$是中线,$BC = 3$,$AC = 2$,则$\triangle BCD与\triangle ACD$的周长差为______
1
;(2)若$CD$是高,$∠ABC = 62^{\circ}$,求$∠BOC$的度数;
∵ BE 是$∠ABC$的平分线,$∠ABC = 62^{\circ}$,∴$∠ABE = \frac{1}{2}∠ABC = \frac{1}{2}×62^{\circ} = 31^{\circ}$.∵ CD 是$△ABC$的高,∴$∠CDB = 90^{\circ}$.∴$∠BOC = ∠CDB + ∠ABE = 90^{\circ} + 31^{\circ} = 121^{\circ}$.
(3)若$CD$是角平分线,$∠A = 78^{\circ}$,求$∠BOC$的度数。
在$△ABC$中,$∠A = 78^{\circ}$,∴$∠ABC + ∠ACB = 180^{\circ} - ∠A = 102^{\circ}$.∵ BE 是$∠ABC$的平分线,CD 是$∠ACB$的平分线,∴$∠OBC = \frac{1}{2}∠ABC$,$∠OCB = \frac{1}{2}∠ACB$.∴$∠OBC + ∠OCB = \frac{1}{2}(∠ABC + ∠ACB) = \frac{1}{2}×102^{\circ} = 51^{\circ}$.∴$∠BOC = 180^{\circ} - (∠OBC + ∠OCB) = 180^{\circ} - 51^{\circ} = 129^{\circ}$.
答案
4.(1)1
(2)∵ BE 是$∠ABC$的平分线,$∠ABC = 62^{\circ}$,
∴$∠ABE = \frac{1}{2}∠ABC = \frac{1}{2}×62^{\circ} = 31^{\circ}$.
∵ CD 是$△ABC$的高,∴$∠CDB = 90^{\circ}$.
∴$∠BOC = ∠CDB + ∠ABE = 90^{\circ} + 31^{\circ} = 121^{\circ}$.
(3)在$△ABC$中,$∠A = 78^{\circ}$,
∴$∠ABC + ∠ACB = 180^{\circ} - ∠A = 102^{\circ}$.
∵ BE 是$∠ABC$的平分线,CD 是$∠ACB$的平分线,
∴$∠OBC = \frac{1}{2}∠ABC$,$∠OCB = \frac{1}{2}∠ACB$.
∴$∠OBC + ∠OCB = \frac{1}{2}(∠ABC + ∠ACB) = \frac{1}{2}×102^{\circ} = 51^{\circ}$.
∴$∠BOC = 180^{\circ} - (∠OBC + ∠OCB) = 180^{\circ} - 51^{\circ} = 129^{\circ}$.
(2)∵ BE 是$∠ABC$的平分线,$∠ABC = 62^{\circ}$,
∴$∠ABE = \frac{1}{2}∠ABC = \frac{1}{2}×62^{\circ} = 31^{\circ}$.
∵ CD 是$△ABC$的高,∴$∠CDB = 90^{\circ}$.
∴$∠BOC = ∠CDB + ∠ABE = 90^{\circ} + 31^{\circ} = 121^{\circ}$.
(3)在$△ABC$中,$∠A = 78^{\circ}$,
∴$∠ABC + ∠ACB = 180^{\circ} - ∠A = 102^{\circ}$.
∵ BE 是$∠ABC$的平分线,CD 是$∠ACB$的平分线,
∴$∠OBC = \frac{1}{2}∠ABC$,$∠OCB = \frac{1}{2}∠ACB$.
∴$∠OBC + ∠OCB = \frac{1}{2}(∠ABC + ∠ACB) = \frac{1}{2}×102^{\circ} = 51^{\circ}$.
∴$∠BOC = 180^{\circ} - (∠OBC + ∠OCB) = 180^{\circ} - 51^{\circ} = 129^{\circ}$.
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