a) Ta có:
$CD//Ax \, (gt)$
⇒ $\widehat{ACD} = \widehat{CAx}$ (so le trong)
$\widehat{BAx} = \widehat{ADC}$ (đồng vị)
mà $\widehat{CAx} = \widehat{BAx} = \dfrac{\widehat{BAC}}{2} \, (gt)$
⇒ $\widehat{BAx} = \widehat{ADC} = \widehat{ACD}$
b) Ta có: $\widehat{DAy} = \widehat{CAy} = \dfrac{\widehat{DAC}}{2} \, (gt)$
$\widehat{CAx} = \widehat{BAx} = \dfrac{\widehat{BAC}}{2} \, (gt)$
⇒ $\widehat{xAy} = \widehat{CAy} + \widehat{CAx} = \dfrac{\widehat{DAC}}{2} + \dfrac{\widehat{BAC}}{2} = \dfrac{\widehat{BAD}}{2} = \dfrac{180^o}{2} = 90^o$
c) Gọi $H$ là giao điểm của $Ay$ và $CD$ ta có:
$CD//Ax$
⇒ $\widehat{CHy} = \widehat{xAH}$ (đồng vị)
mà $\widehat{xAH} = \widehat{xAy} = 90^o$
nên $\widehat{CHy} = 90^o$
hay $Ay\perp CD$