Ta có: $AB//CD⇒AM//CD(M∈AB)$
$⇒\widehat{AMD}=\widehat{MDC}(1)$
$AM=AD⇒ΔAMD$ cân tại $A$
$⇒\widehat{AMD}=\widehat{ADM}(2)$
Từ (1)(2)$⇒\widehat{MDC}=\widehat{ADM}$
$⇒DM$ là tia phân giác $\widehat{D}$
Do $AB//CD⇒\widehat{MBC}+\widehat{BCD}=180^o$ (2 góc trong cùng phía)
$⇒\widehat{MBC}+80^o=180^o$
$⇒\widehat{MBC}=100^o$