Hạ AM;BN⊥DC cắt EF lần lượt tại I;K
Ta có:
$\frac{EI}{DM}$ = $\frac{AE}{AD}$ =$\frac{p}{p+q}$ ⇔$EI=\frac{p}{p+q} DM$
$\frac{KF}{NC}$ = $\frac{BF}{BC}$ =$\frac{p}{p+q}$ ⇔$KF=\frac{p}{p+q} NC$
Suy ra: $EF=EI+IK+KF=\frac{p}{p+q} DM+AB+\frac{p}{p+q} NC$
⇔$EF=\frac{p}{p+q} (DM+NC)+AB$
⇔$EF=\frac{p}{p+q}(DC-AB)+AB$
⇔$EF=\frac{p}{p+q}DC+\frac{q}{p+q}AB$ (Đpcm)