b) Xét $\triangle$BMC và $\triangle$DMA có:
AM = MC
BM = MD
$\widehat{AMD}$ = $\widehat{CMB}$
Vậy $\triangle$BMC = $\triangle$DMA (c.g.c)
$\rightarrow$ $\widehat{MAD}$ = $\widehat{MCB}$ $\rightarrow$ AD//BC
Và AH $\bot$ BC $\rightarrow$ AH $\bot$ AD (đpcm)
c) Xét $\triangle$AMB và $\triangle$CMD có:
AM = MC
BM = MD
$\widehat{AMB}$ = $\widehat{CMD}$ ( 2 góc đối đỉnh)
Vậy $\triangle$AMB = $\triangle$CMD (c.g.c)
$\rightarrow$ $\widehat{ADM}$ = $\widehat{MDC}$
Ta có:
$\widehat{ABC}$ =$\widehat{ABM}$ + $\widehat{MBC}$
$\widehat{CDA}$ = $\widehat{CDM}$ + $\widehat{MDA}$
$\Longrightarrow$ $\widehat{ABC}$ = $\widehat{CDA}$ (đpcm)