第156页 - 苏科版八年级（初二）数学学习与评价答案（上下册）

证明：$ $在$△ ADC $和$△A_1\ \mathrm {D}_1\ \mathrm {C}_1 $中

$ \begin {cases}{A D=A_1\ \mathrm {D}_1}\\{D C=D_1\ \mathrm {C}_1 }\\{A C=A_1\ \mathrm {C}_1}\end {cases}$

∴$△A D C ≌△A_1\ \mathrm {D}_1\ \mathrm {C}_1(S S S) $

又在$△ABC $和$ △A_1\ \mathrm {B}_1\ \mathrm {C}_1 $中

$\begin {cases}{A B=A_1\ \mathrm {B}_1}\\{B C=B_1\ \mathrm {C}_1 }\\{A C=A_1\ \mathrm {C}_1}\end {cases}$

∴$△A B C≌△A_1\ \mathrm {B}_1\ \mathrm {C}_1(S S S) $

∴$∠B =∠B_1$，$ ∠B A D=∠B_1\ \mathrm {A}_1\ \mathrm {D}_1 ∠D=∠D_1 $，

$∠BCD=∠B_1\ \mathrm {C}_1\ \mathrm {D}_1 $

∴四边形$ ABCD $与四边形$ A_1\ \mathrm {B}_1\ \mathrm {C}_1\ \mathrm {D}_1 $全等

解$ ∶ ∠A=∠A_1$

理由：连接$ B D $、$ B_1\ \mathrm {D}_1$

在$ △ABD $和$△A_1\ \mathrm {B}_1\ \mathrm {D}_1 $中

$\begin {cases}{A D=A_1\ \mathrm {D}_1}\\{∠A=∠A_1 }\\{A B=A_1\ \mathrm {B}_1}\end {cases}$

∴$△A B D ≌△ A_1\ \mathrm {B}_1\ \mathrm {D}_1(S A S)$

∴$B D=B_1\ \mathrm {D}_1$

由$ (1 ) $可知四边形$ ABCD$与四边形$ A_1\ \mathrm {B}_1\ \mathrm {C}_1\ \mathrm {D}_1 $全等$.$

解：条件$ A C=A_1\ \mathrm {C}_1$，$ A D=A_1\ \mathrm {D}_1$