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C

B

2.

$AC=DF.$

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$證明: ∵ ∠BCE = ∠ACD= 90°,\ $

$∴∠BCA = ∠ECD.\ $

$在△BCA 和△ECD 中，$

$\begin{cases}{∠BCA=∠ECD}\\{∠BAC=∠D}\\{BC=CE}\end{cases}$

$∴△BCA≌△ECD$

$∴AC=CD$

$證明: (1)由題意得: A C=B C, \angle A C B=90^{\circ},\ $

$A D \perp D E, B E \perp D E,\ $

$\therefore \angle A D C=\angle C E B=90^{\circ},\ $

$\therefore \angle A C D+\angle B C E=90^{\circ}, \angle A C D+\angle D A C=90^{\circ},$

$\ \therefore \angle B C E=\angle D A C,$

$在\triangle A D C 和 \triangle C E B中,\ \ $

$\left\{\begin{array}{l}\angle A D C=\angle C E B \\ \angle D A C=\angle B C E \text {, } \\ A C=B C\end{array}\right.$

$\therefore \triangle A D C \cong \triangle C E B(\mathrm{AAS}) $

$解:(2)由題意得:\ 一塊墻磚的厚度為a,$

$\therefore A D=4\ \mathrm {a}, B E=3\ \mathrm {a},$

$由(1)得:\triangle A D C \cong \triangle C E B,\ $

$\therefore D C=B E=3\ \mathrm {a}, A D=C E=4\ \mathrm {a},\ $

$\therefore D C+C E=B E+A D=7\ \mathrm {a}=35,\ $

$\therefore a=5,$

$∴砌墻磚塊的厚度a為5\ \mathrm {cm}.$

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