a,
Ta có: $\dfrac{AB}{PM}=\dfrac{45}{12}=\dfrac{15}{4}$
$\dfrac{BC}{MN}=\dfrac{75}{20}=\dfrac{15}{4}$
⇒ $\dfrac{AB}{PM}=\dfrac{BC}{MN}$
Xét $ΔMNP$ có:
$MP^2+NP^2=12^2+16^2=400=20^2$
Mà $MN^2=20^2=400$
⇒$MP^2+NP^2=MN^2$
⇒$ΔMNP$ vuông tại$P$
Xét $ΔABC$ và $ΔPMN$ có:
$\widehat{BAC}=\widehat{PMN}=90^o$
$\dfrac{AB}{PM}=\dfrac{BC}{MN}(cmt)$
⇒$ΔABC$$\sim$$ΔPMN(c.g,c)$
b,Có:
$ΔABC$$\sim$$ΔPMN(c.g,c)$
⇒$\dfrac{S_{PMN}}{S_{ABC}}=\dfrac{PM}{AB}^2=\dfrac{4}{15}^2=\dfrac{16}{225}$
Do $\dfrac{AB}{PM}=\dfrac{45}{12}=\dfrac{15}{4}$