25. （2023·达州）（1）如图1，在矩形$ABCD$的$AB$边上取一点$E$，将$△ ADE$沿$DE$翻折，使点$A$落在$BC$上$A'$处，若$AB = 6$，$BC = 10$，求$\frac{AE}{EB}$的值；
（2）如图2，在矩形$ABCD$的$BC$边上取一点$E$，将四边形$ABED$沿$DE$翻折，使点$B$落在$DC$的延长线上$B'$处，若$BC· CE = 24$，$AB = 6$，求$BE$的值；
（3）如图3，在$△ ABC$中，$∠ BAC = 45^{\circ}$，$AD⊥ BC$，垂足为点$D$，$AD = 10$，$AE = 6$，过点$E$作$EF⊥ AD$交$AC$于点$F$，连接$DF$，且满足$∠ DFE = 2∠ DAC$，直接写出$BD + \frac{5}{3}EF$的值.

25. (1) $ \because $四边形$ ABCD $是矩形，$ \therefore A D = B C = 10 $，$ C D = A B = 6 $，$ ∠ A = ∠ B = ∠ C = 90 ^ { \circ } $。由翻折性质得$ A D = A ^ { \prime } D = 10 $，$ A E = A ^ { \prime } E $，在$ \mathrm { Rt } △ A ^ { \prime } C D $中，$ A ^ { \prime } C = \sqrt { A ^ { \prime } D ^ { 2 } - C D ^ { 2 } } = \sqrt { 10 ^ { 2 } - 6 ^ { 2 } } = 8 $，$ \therefore A ^ { \prime } B = B C - A ^ { \prime } C = 2 $。设$ A E = A ^ { \prime } E = x $，则$ B E = A B - A E = 6 - x $，在$ \mathrm { Rt } △ A ^ { \prime } B E $中，由勾股定理得$ B E ^ { 2 } + A ^ { \prime } B ^ { 2 } = A ^ { \prime } E ^ { 2 } $，$ \therefore ( 6 - x ) ^ { 2 } + 2 ^ { 2 } = x ^ { 2 } $。解得$ x = \dfrac { 10 } { 3 } $，$ \therefore A E = \dfrac { 10 } { 3 } $，$ B E = 6 - \dfrac { 10 } { 3 } = \dfrac { 8 } { 3 } $，$ \therefore \dfrac { A E } { E B } = \dfrac { \dfrac { 10 } { 3 } } { \dfrac { 8 } { 3 } } = \dfrac { 5 } { 4 } $；(2) $ \because $四边形$ ABCD $是矩形，$ \therefore C D = A B = 6 $，$ A D = B C $，$ ∠ A = ∠ B = ∠ B C D = 90 ^ { \circ } $。由翻折性质得，$ A ^ { \prime } B ^ { \prime } = A B = 6 $，$ A ^ { \prime } D = A D $，$ ∠ D A ^ { \prime } B ^ { \prime } = ∠ A B E = ∠ B C D = 90 ^ { \circ } $，$ \therefore ∠ E B ^ { \prime } C + ∠ A ^ { \prime } B ^ { \prime } D = 90 ^ { \circ } = ∠ A ^ { \prime } B ^ { \prime } D + ∠ B ^ { \prime } D A ^ { \prime } $，$ \therefore ∠ E B ^ { \prime } C = ∠ B ^ { \prime } D A ^ { \prime } $，$ \therefore △ E B ^ { \prime } C ∽ △ B ^ { \prime } D A ^ { \prime } $，$ \therefore \dfrac { C E } { A ^ { \prime } B ^ { \prime } } = \dfrac { B ^ { \prime } C } { A ^ { \prime } D } $，即$ \dfrac { C E } { 6 } = \dfrac { B ^ { \prime } C } { B C } $。又$ B C · C E = 24 $，$ \therefore B ^ { \prime } C = \dfrac { B C · C E } { 6 } = \dfrac { 24 } { 6 } = 4 $，$ \therefore B ^ { \prime } D = B ^ { \prime } C + C D = 10 $。在$ \mathrm { Rt } △ A ^ { \prime } B ^ { \prime } D $中，$ A ^ { \prime } D = \sqrt { B ^ { \prime } D ^ { 2 } - A ^ { \prime } B ^ { \prime 2 } } = 8 $，$ \therefore B C = A D = A ^ { \prime } D = 8 $，则$ C E = 3 $，$ \therefore B E = B C - C E = 8 - 3 = 5 $；(3) $ \because A D ⊥ B C $，$ E F ⊥ A D $，$ \therefore E F // B C $，$ \therefore △ A E F ∽ △ A D C $。$ \because A D = 10 $，$ A E = 6 $，$ \therefore \dfrac { E F } { C D } = \dfrac { A E } { A D } = \dfrac { 6 } { 10 } = \dfrac { 3 } { 5 } $，$ \therefore C D = \dfrac { 5 } { 3 } E F $。则$ B D + \dfrac { 5 } { 3 } E F = B D + C D = B C $。设$ E F = 3 k $，$ C D = 5 k $，过点$ D $作$ D H ⊥ A C $于$ H $，则$ ∠ C H D = ∠ A D C = 90 ^ { \circ } $，$ \therefore ∠ C D H = ∠ D A C = 90 ^ { \circ } - ∠ C $。$ \because E F // B C $，$ \therefore ∠ C D F = ∠ D F E = 2 ∠ D A C = 2 ∠ C D H $，$ \therefore ∠ C D H = ∠ F D H $。又$ \because D H = D H $，$ ∠ C H D = ∠ F H D = 90 ^ { \circ } $，$ \therefore △ C H D ≌ △ F H D ( \mathrm { ASA } ) $，$ \therefore D F = C D = 5 k $。在$ \mathrm { Rt } △ E F D $中，由勾股定理得$ E F ^ { 2 } + D E ^ { 2 } = D F ^ { 2 } $，$ \therefore ( 3 k ) ^ { 2 } + 4 ^ { 2 } = ( 5 k ) ^ { 2 } $，解得$ k = 1 $，$ \therefore E F = 3 $，$ D F = C D = 5 $。在$ \mathrm { Rt } △ A D C $中，$ A C = \sqrt { A D ^ { 2 } + C D ^ { 2 } } = \sqrt { 10 ^ { 2 } + 5 ^ { 2 } } = 5 \sqrt { 5 } $。过$ B $作$ B G ⊥ A C $于$ G $，则$ ∠ B G A = ∠ B G C = ∠ C H D = 90 ^ { \circ } $，$ \therefore B G // D H $，$ \therefore ∠ C B G = ∠ C D H = ∠ D A C $，$ \therefore \dfrac { C D } { A C } = \dfrac { 5 } { 5 \sqrt { 5 } } = \dfrac { \sqrt { 5 } } { 5 } $，$ \dfrac { A D } { A C } = \dfrac { 10 } { 5 \sqrt { 5 } } = \dfrac { 2 \sqrt { 5 } } { 5 } $。$ \because ∠ B A C = 45 ^ { \circ } $，$ ∠ A G B = 90 ^ { \circ } $，$ \therefore ∠ A B G = 90 ^ { \circ } - ∠ B A C = 45 ^ { \circ } = ∠ B A C $，$ \therefore A G = B G $。在$ \mathrm { Rt } △ B C G $中，$ B G = \dfrac { 2 \sqrt { 5 } } { 5 } B C $，$ C G = \dfrac { \sqrt { 5 } } { 5 } B C $。$ \because A G + C G = B G + C G = A C $，$ \therefore \dfrac { 2 \sqrt { 5 } } { 5 } B C + \dfrac { \sqrt { 5 } } { 5 } B C = 5 \sqrt { 5 } $，$ \therefore B C = \dfrac { 25 } { 3 } $，$ \therefore B D + \dfrac { 5 } { 3 } E F = B C = \dfrac { 25 } { 3 } $。