a: ta có: BN=AD
mà AD=BC
nên BN=BC
=>ΔBNC cân tại B
Ta có: AB=DM
mà AB=DC
nên DM=DC
=>ΔDMC cân tại D
b: Xét ΔCBN và ΔMDC có
\(\dfrac{CB}{MD}=\dfrac{BN}{DC}\)
\(\widehat{CBN}=\widehat{MDC}\left(=\widehat{BCD}\right)\)
Do đó: ΔCBN~ΔMDC
c: ΔCBN~ΔMDC
=>\(\widehat{BCN}=\widehat{DMC}\)
mà \(\widehat{DMC}=\dfrac{180^0-\widehat{MDC}}{2}=\dfrac{\widehat{ADC}}{2}\)
nên \(\widehat{BCN}=\dfrac{\widehat{ADC}}{2}\)
\(\widehat{BCN}+\widehat{DCM}+\widehat{DCB}=\dfrac{\widehat{ADC}}{2}+\dfrac{180^0-\widehat{MDC}}{2}+\widehat{DCB}\)
\(=\dfrac{\widehat{ADC}}{2}+\dfrac{\widehat{ADC}}{2}+\widehat{DCB}=\widehat{ADC}+\widehat{DCB}=180^0\)
=>M,C,N thẳng hàng
