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Giải thích các bước giải:

1.Xét $\Delta DAB,\Delta ABC$ có:

Chung $AB$

$\hat A=\hat B$

$AD=BC$

$\to \Delta ABD=\Delta BAC(c.g.c)$

$\to BD=CA$

2.Xét $\Delta ACD,\Delta BDC$ có:

Chung $CD$

$AD=BC$

$AC=BD$

$\to \Delta ADC=\Delta BCD(c.c.c)$

$\to \widehat{ADC}=\widehat{BCD}$

3.Từ câu 1 $\to \widehat{DAB}=\widehat{ABC}$

Ta có:

$\widehat{ABC}+\widehat{BCD}+\widehat{CDA}+\widehat{DAB}=360^o$

Mà $\widehat{DAB}=\widehat{ABC} , \widehat{ADC}=\widehat{DCB}$

$\to2(\widehat{DAB}+\widehat{ADC})=360^o$

$\to \widehat{DAB}+\widehat{ADC}=180^o$

$\to AB//CD$

Do $AC=BD$

$\to ABCD$ là hình thang cân