$\widehat{BAC}+\widehat{ACD}=180^\circ$
mà $\widehat{BAC}=130^\circ$
$→\widehat{ACD}=180^\circ-130^\circ=50^\circ$
Xét $ΔACD$:
$\widehat{ACD}+\widehat{ADC}+\widehat{CAD}=180^\circ$ (tổng 3 góc trong 1 Δ)
$→\widehat{CAD}=180^\circ-\widehat{ACD}-\widehat{ADC}=180^\circ-40^\circ-50^\circ=90^\circ$
$→AC⊥AD$ (ĐPCM)