Đáp án:

Bài 1:

a) $\triangle AEB=\triangle AMB$

b) $\triangle BAN=\triangle BAC$

c) $\triangle NEA=\triangle CMA$

Bài 2:

a) $\triangle EAC=\triangle ENC$

b) $\triangle ECM=\triangle ECH$

c) \triangle MAC=\triangle HNC$

Giải thích các bước giải:

Bài 1:

a)

Xét $\triangle AEB$ và $\triangle AMB$:

$\widehat{AEB}=\widehat{AMB}\,\,\,(=90^o)$

$AB$: chung

$\widehat{ABE}=\widehat{ABM}$ (gt)

$\to\triangle AEB=\triangle AMB$ (ch - gn)

b)

Xét $\triangle BAN$ và $\triangle BAC$:

$\widehat{BAN}=\widehat{BAC}\,\,\,(=90^o)$

$BA$: chung

$\widehat{ABN}=\widehat{ABC}$ (gt)

$\to\triangle BAN=\triangle BAC$ (cgv - gn)

$\to NA=CA$ (2 cạnh tương ứng)

$\widehat{ANB}=\widehat{ACB}$ (2 góc tương ứng)

c)

Xét $\triangle NEA$ và $\triangle CMA$:

$\widehat{NEA}=\widehat{CMA}\,\,\,(=90^o)$

$NA=CA$ (cmt)

$\widehat{ANE}=\widehat{ACM}\,\,\,(\widehat{ANB}=\widehat{ACB})$

$\to\triangle NEA=\triangle CMA$ (ch - gn)

Bài 2:

a)

Xét $\triangle EAC$ và $\triangle ENC$:

$\widehat{EAC}=\widehat{ENC}\,\,\,(=90^o)$

$EA=EN$ (gt)

$EC$: chung

$\to\triangle EAC=\triangle ENC$ (ch - cgv)

$\to\widehat{AEC}=\widehat{NEC}$ (2 góc tương ứng)

b)

Xét $\triangle ECM$ và $\triangle ECH$:

$\widehat{ECM}=\widehat{ECH}\,\,\,(=90^o)$

$EC$: chung

$\widehat{MEC}=\widehat{HEC}\,\,\,(\widehat{AEC}=\widehat{NEC})$

$\to\triangle ECM=\triangle ECH$ (cgv - gn)

$\to CM=CH$ (2 cạnh tương ứng)

$\widehat{EMC}=\widehat{EHC}$ (2 góc tương ứng)

c)

Xét $\triangle MAC$ và $\triangle HNC$:

$\widehat{MAC}=\widehat{HNC}\,\,\,(=90^o)$

$CM=CH$ (cmt)

$\widehat{AMC}=\widehat{NHC}\,\,\,(\widehat{EMC}=\widehat{EHC})$

$\to\triangle MAC=\triangle HNC$ (ch - gn)