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7. 感知:如图①,$AD平分\angle BAC$,$\angle B + \angle C = 180^{\circ}$,$\angle B = 90^{\circ}$. 易知:$DB = DC$. (不需证明)
探究:如图②,$AD平分\angle BAC$,$\angle ABD + \angle ACD = 180^{\circ}$,$\angle ABD＜90^{\circ}$. 求证:$DB = DC$.
应用:如图③,在四边形$ABDC$中,$\angle B = 45^{\circ}$,$\angle C = 135^{\circ}$,$DB = DC$,$DE\perp AB于E$,求证:$AB - AC = 2BE$.
探究:如图②,$AD平分\angle BAC$,$\angle ABD + \angle ACD = 180^{\circ}$,$\angle ABD＜90^{\circ}$. 求证:$DB = DC$.
应用:如图③,在四边形$ABDC$中,$\angle B = 45^{\circ}$,$\angle C = 135^{\circ}$,$DB = DC$,$DE\perp AB于E$,求证:$AB - AC = 2BE$.
答案
探究: 作 $ D E \perp A B $ 于 $ E $, $ D F \perp A C $, 交 $ A C $ 的延长线于 $ F $, 如图①所示. 在 $ \triangle D F A $ 和 $ \triangle D E A $ 中, $ \left\{ \begin{array} { l } { \angle D A F = \angle D A E, } \\ { \angle A F D = \angle A E D = 90 ^ { \circ }, } \\ { D A = D A, } \end{array} \right. $ $ \therefore \triangle D F A \cong \triangle D E A ( \mathrm { AAS } ) $, $ \therefore D F = D E $. $ \because \angle A B D + \angle A C D = 180 ^ { \circ } $, $ \angle A C D + \angle F C D = 180 ^ { \circ } $, $ \therefore \angle A B D = \angle F C D $. 在 $ \triangle D F C $ 和 $ \triangle D E B $ 中, $ \left\{ \begin{array} { l } { \angle D F C = \angle D E B, } \\ { \angle F C D = \angle E B D, } \\ { D F = D E, } \end{array} \right. $ $ \therefore \triangle D F C \cong \triangle D E B ( \mathrm { AAS } ) $, $ \therefore D B = D C $.
应用: 连接 $ A D $, 作 $ D F \perp A C $, 交 $ A C $ 的延长线于 $ F $, 如图②所示. $ \because \angle A C D = 135 ^ { \circ } $, $ \therefore \angle F C D = 180 ^ { \circ } - \angle A C D = 45 ^ { \circ } $. $ \because \angle B = 45 ^ { \circ } $, $ \therefore \angle F C D = \angle B $. 在 $ \triangle D F C $ 和 $ \triangle D E B $ 中, $ \left\{ \begin{array} { l } { \angle D F C = \angle D E B = 90 ^ { \circ }, } \\ { \angle F C D = \angle B, } \\ { D C = D B, } \end{array} \right. $ $ \therefore \triangle D F C \cong \triangle D E B ( \mathrm { AAS } ) $, $ \therefore D F = D E $, $ C F = B E $. 在 $ \mathrm { Rt } \triangle A D F $ 和 $ \mathrm { Rt } \triangle A D E $ 中, $ \left\{ \begin{array} { l } { A D = A D, } \\ { D F = D E, } \end{array} \right. $ $ \therefore \mathrm { Rt } \triangle A D F \cong \mathrm { Rt } \triangle A D E ( \mathrm { HL } ) $, $ \therefore A F = A E $, $ \therefore A B = A E + B E = A C + C F + B E = A C + 2 B E $, $ \therefore A B - A C = 2 B E $.
8. 如图,在$\triangle ABC$中,$\angle BAC = 60^{\circ}$,$\angle C = 40^{\circ}$,$AP平分\angle BAC交BC于点P$,$BQ平分\angle ABC交AC于点Q$,求证:$AB + BP = BQ + AQ$.
答案
如图, 过点 $ P $ 作 $ B Q $ 的平行线交 $ A C $ 于点 $ D $. $ \because B Q $ 平分 $ \angle A B C $ 且 $ \angle A B C = 180 ^ { \circ } - \angle B A C - \angle C = 80 ^ { \circ } $, $ \therefore \angle C B Q = \frac { 1 } { 2 } \angle A B C = \frac { 1 } { 2 } \times 80 ^ { \circ } = 40 ^ { \circ } $, $ \therefore \angle C B Q = \angle A C B $. 过点 $ Q $ 作 $ Q F \perp B C $ 于点 $ F $, 易证 $ \triangle B Q F \cong \triangle C Q F $, $ \therefore B Q = C Q $, $ \therefore B Q + A Q = C Q + A Q = A C $ ①. $ \because P D // B Q $, $ \therefore \angle C P D = \angle C B Q = 40 ^ { \circ } $, $ \therefore \angle C P D = \angle A C B = 40 ^ { \circ } $, 易得 $ P D = C D $. 又 $ \angle A D P = \angle C P D + \angle A C B = 40 ^ { \circ } + 40 ^ { \circ } = 80 ^ { \circ } $, 且 $ \angle A B C = 80 ^ { \circ } $, $ \therefore \angle A B C = \angle A D P $. $ \because A P $ 平分 $ \angle B A C $, $ \therefore \angle B A P = \angle C A P $. 在 $ \triangle A B P $ 与 $ \triangle A D P $ 中, $ \left\{ \begin{array} { l } { \angle A B P = \angle A D P, } \\ { \angle B A P = \angle D A P, } \\ { A P = A P, } \end{array} \right. $ $ \therefore \triangle A B P \cong \triangle A D P ( \mathrm { AAS } ) $, $ \therefore A B = A D $, $ B P = P D $, $ \therefore A B + B P = A D + P D = A D + C D = A C $ ②. 由 ①② 可得, $ A B + B P = B Q + A Q $.
9. 如图,在$Rt\triangle ABC$中,$\angle BAC = 90^{\circ}$,点$D$,$E在边BC$上,$\angle CAE = \angle B$,$E是CD$的中点,且$AD平分\angle BAE$,试问:$BD与AC$相等吗? 请说说你的理由.
答案
$ B D = A C $. 理由: 如图, 由于 $ A D $ 平分 $ \angle B A E $, 所以可将 $ \triangle A B D $ 沿 $ A D $ 所在直线翻折到 $ \triangle A F D $ 的位置, 则 $ \triangle A B D \cong \triangle A F D $, $ \therefore \angle F = \angle B $, $ B D = F D $. 又 $ \angle C A E = \angle B $, $ \therefore \angle C A E = \angle F $. 又 $ \angle A E C = \angle F E D $, 且 $ E $ 是 $ C D $ 的中点, $ \therefore \triangle A C E \cong \triangle F D E $, $ \therefore D F = A C $, $ \therefore B D = A C $.
10. 如图,在正方形$ABCD$中,点$E是BC$上一点,点$F是DC$上一点,$\angle EAF = 45^{\circ}$.
(1)如图①,若$BE = DF = 1$,求$EF$的长.
(2)如图②,求证:$BE + DF = EF$.
(3)如图③,点$E为CB$延长线上一点,点$F为DC$延长线上一点,$\angle EAF = 45^{\circ}$. 请直接写出线段$BE$,$DF$,$EF$的数量关系.
(1)如图①,若$BE = DF = 1$,求$EF$的长.
(2)如图②,求证:$BE + DF = EF$.
(3)如图③,点$E为CB$延长线上一点,点$F为DC$延长线上一点,$\angle EAF = 45^{\circ}$. 请直接写出线段$BE$,$DF$,$EF$的数量关系.
答案
(1) 如图①, 将 $ \triangle A D F $ 绕着点 $ A $ 按顺时针方向旋转 $ 90 ^ { \circ } $, 得 $ \triangle A B H $, $ \therefore D F = B E = B H = 1 $. $ \because $ 四边形 $ A B C D $ 是正方形, $ \therefore A D = A B = B C = C D $, $ \angle A B C = \angle A D F = \angle C = 90 ^ { \circ } $. 由旋转可得 $ A F = A H $, $ \angle D A F = \angle B A H $, 又 $ \because \angle E A F = 45 ^ { \circ } $, $ \therefore \angle D A F + \angle B A E = 45 ^ { \circ } = \angle B A H + \angle B A E = \angle H A E = \angle E A F $. 又 $ \because A E = A E $, $ \therefore \triangle A E F \cong \triangle A E H ( \mathrm { SAS } ) $, $ \therefore H E = E F = 2 $.
(2) 如图②, 将 $ \triangle A D F $ 绕着点 $ A $ 按顺时针方向旋转 $ 90 ^ { \circ } $, 得 $ \triangle A B F ^ { \prime } $, 则 $ \angle A B F ^ { \prime } = \angle D $, $ A F = A F ^ { \prime } $, $ B F ^ { \prime } = D F $, $ \angle B A F ^ { \prime } = \angle D A F $. $ \because $ 四边形 $ A B C D $ 是正方形, $ \therefore \angle D = \angle A B C = 90 ^ { \circ } $, $ \therefore \angle A B F ^ { \prime } = 90 ^ { \circ } $, $ \therefore \angle F ^ { \prime } B C = 180 ^ { \circ } $, $ \therefore F ^ { \prime } $, $ B $, $ E $ 在同一直线上. $ \because \angle E A F = 45 ^ { \circ } $, $ \therefore \angle D A F + \angle B A E = 45 ^ { \circ } = \angle F ^ { \prime } A B + \angle B A E = \angle F ^ { \prime } A E = \angle E A F $. 又 $ \because A E = A E $, $ \therefore \triangle A F ^ { \prime } E \cong \triangle A F E ( \mathrm { SAS } ) $, $ \therefore E F = E F ^ { \prime } = B E + B F ^ { \prime } = B E + D F $, 即 $ B E + D F = E F $.
(3) $ E F = D F - B E $. 解析: 如图③, 将 $ \triangle A B E $ 绕着点 $ A $ 按逆时针方向旋转 $ 90 ^ { \circ } $, $ \because $ 四边形 $ A B C D $ 是正方形, $ \therefore A B $ 与 $ A D $ 重合, 点 $ E $ 落在 $ C D $ 上的点 $ H $ 处, 得 $ \triangle A D H $, $ \therefore \triangle A D H \cong \triangle A B E $, $ \therefore \angle D A H = \angle B A E $, $ D H = B E $, $ A E = A H $. $ \because \angle E A F = 45 ^ { \circ } $, $ \therefore \angle B A E + \angle B A F = 45 ^ { \circ } = \angle D A H + \angle B A F $, $ \therefore \angle F A H = 90 ^ { \circ } - \angle D A H - \angle B A F = 45 ^ { \circ } = \angle E A F $. 又 $ \because A F = A F $, $ A E = A H $, $ \therefore \triangle A E F \cong \triangle A H F ( \mathrm { SAS } ) $, $ \therefore E F = F H $, $ \therefore E F = F H = D F - D H = D F - B E $.
