4. (1)如图①,在$\triangle ABC$中,$AB = AC$,$\angle BAC = 90^{\circ}$,过点$A作AH\perp BC于H$,求证:$AH = \frac{1}{2}BC$.
(2)在如图②和图③的两张图中,在$\triangle ABC$中,$AB = AC$,且$\angle BAC = 90^{\circ}$,在同一平面内有一点$P$,满足$PC = 1$,$PB = 6$,且$\angle BPC = 90^{\circ}$,请分别求出两图中点$A到BP$的距离.

(1) $ \because A H \perp B C $, $ \angle B A C = 90 ^ { \circ } $, $ \therefore \angle A H C = 90 ^ { \circ } = \angle B A C $, $ \therefore \angle B A H + \angle C A H = 90 ^ { \circ } $, $ \angle B A H + \angle B = 90 ^ { \circ } $, $ \therefore \angle C A H = \angle B $. 在 $ \triangle A B H $ 和 $ \triangle C A H $ 中, $ \left\{ \begin{array} { l } { \angle B = \angle C A H, } \\ { \angle B H A = \angle A H C, } \\ { A B = C A, } \end{array} \right. $ $ \therefore \triangle A B H \cong \triangle C A H ( A A S ) $, $ \therefore B H = A H $, $ A H = C H $, $ \therefore A H = \frac { 1 } { 2 } B C $.
(2) 如图①, 过点 $ A $ 作 $ A H \perp B P $ 于点 $ H $, 连接 $ A P $, 在 $ B P $ 上取一点 $ D $ 使得 $ B D = P C $, $ \therefore D P = B P - B D = 6 - 1 = 5 $. 设 $ A C $ 与 $ B P $ 交于点 $ E $. $ \because \angle B A C = \angle B P C = 90 ^ { \circ } $ 且 $ \angle A E B = \angle P E C $, $ \therefore \angle A B D = \angle A C P $. 又 $ \because A B = A C $, $ B D = C P $, $ \therefore \triangle A D B \cong \triangle A P C ( S A S ) $, $ \therefore A D = A P $, $ \angle B A D = \angle P A C $, $ \therefore \angle D A P = \angle D A E + \angle P A C = \angle B A D + \angle D A E = \angle B A C = 90 ^ { \circ } $. $ \because A H \perp D P $, $ \therefore A H = \frac { 1 } { 2 } D P = \frac { 5 } { 2 } $.

如图②, 过点 $ A $ 作 $ A H \perp B P $ 于点 $ H $, 连接 $ A P $, 在 $ P B $ 的延长线上取一点 $ D $ 使得 $ B D = P C $, $ \therefore D P = B P + B D = 6 + 1 = 7 $. $ \because \angle B A C = \angle B P C = 90 ^ { \circ } $, $ \therefore \angle A B P + \angle A C P = 180 ^ { \circ } $. $ \because \angle A B P + \angle A B D = 180 ^ { \circ } $, $ \therefore \angle A B D = \angle A C P $. 又 $ \because A B = A C $, $ B D = C P $, $ \therefore \triangle A D B \cong \triangle A P C ( S A S ) $, $ \therefore A D = A P $, $ \angle B A D = \angle C A P $, $ \therefore \angle D A P = \angle D A B + \angle B A P = \angle C A P + \angle B A P = \angle B A C = 90 ^ { \circ } $. $ \because A H \perp D P $, $ \therefore A H = \frac { 1 } { 2 } D P = \frac { 7 } { 2 } $.

5. (2025·毕节校级月考)如图,已知$AD// BC$,$\angle PAB的平分线与\angle CBA的平分线相交于点E$,$CE的延长线交AP于点D$. 求证:$AD + BC = AB$.

在 $ A B $ 上截取 $ A F = A D $, 连接 $ E F $. $ \because A E $ 平分 $ \angle P A B $, $ \therefore \angle D A E = \angle F A E $. 在 $ \triangle D A E $ 和 $ \triangle F A E $ 中, $ \left\{ \begin{array} { l } { A D = A F, } \\ { \angle D A E = \angle F A E, } \\ { A E = A E, } \end{array} \right. $ $ \therefore \triangle D A E \cong \triangle F A E ( S A S ) $, $ \therefore \angle A F E = \angle A D E $. $ \because A D // B C $, $ \therefore \angle A D E + \angle C = 180 ^ { \circ } $. $ \because \angle A F E + \angle E F B = 180 ^ { \circ } $, $ \therefore \angle E F B = \angle C $. $ \because B E $ 平分 $ \angle A B C $, $ \therefore \angle E B F = \angle E B C $. 在 $ \triangle B E F $ 和 $ \triangle B E C $ 中, $ \left\{ \begin{array} { l } { \angle E F B = \angle C, } \\ { \angle E B F = \angle E B C, } \\ { B E = B E, } \end{array} \right. $ $ \therefore \triangle B E F \cong \triangle B E C ( A A S ) $, $ \therefore B C = B F $, $ \therefore A D + B C = A F + B F = A B $.

6. (2024·北京期末)如图,动点$C与线段AB构成\triangle ABC$,其边长满足$AB = 9$,$CA = 2a + 2$,$CB = 2a - 3$. 点$D在\angle ACB$的平分线上,且$\angle ADC = 90^{\circ}$.
(1)$a$的取值范围是____;
(2)求$\triangle ABD$的面积的最大值.

(1) $ a ＞ \frac { 5 } { 2 } $ 解析: $ \because $ 在 $ \triangle A B C $ 中, $ A C + B C ＞ A B $, $ \therefore 2 a + 2 + 2 a - 3 ＞ 9 $, 解得 $ a ＞ \frac { 5 } { 2 } $. $ \because A C + A B ＞ B C $, $ \therefore 2 a + 2 + 9 ＞ 2 a - 3 $, 恒成立. $ \because B C + A B ＞ A C $, $ \therefore 2 a - 3 + 9 ＞ 2 a + 2 $, 恒成立, $ \therefore a ＞ \frac { 5 } { 2 } $.
(2) 如图, 延长 $ A D $, $ C B $ 交于点 $ E $, $ \because C D $ 为 $ \angle A C B $ 的平分线, $ \therefore \angle A C D = \angle E C D $. 在 $ \triangle A C D $ 和 $ \triangle E C D $ 中, $ \left\{ \begin{array} { l } { \angle A C D = \angle E C D, } \\ { C D = C D, } \\ { \angle A D C = \angle E D C = 90 ^ { \circ }, } \end{array} \right. $ $ \therefore \triangle A C D \cong \triangle E C D ( \mathrm { ASA } ) $, $ \therefore A C = E C = 2 a + 2 $, $ A D = E D $. $ \because C B = 2 a - 3 $, $ \therefore B E = 2 a + 2 - ( 2 a - 3 ) = 5 $. $ \because A D = E D $, $ \therefore S _ { \triangle A B D } : S _ { \triangle A B E } = 1 : 2 $. 当 $ B E \perp A B $ 时, $ \triangle A B E $ 的面积取最大值, 即 $ ( S _ { \triangle A B E } ) _ { \max } = \frac { 1 } { 2 } \times 9 \times 5 = \frac { 45 } { 2 } $, $ \therefore ( S _ { \triangle A B D } ) _ { \max } = \frac { 45 } { 4 } $, $ \therefore \triangle A B D $ 的面积的最大值为 $ \frac { 45 } { 4 } $.