|
$P$即为密克点,有$ \triangle APC\sim \triangle FPD,\frac{PC}{PA}=\frac{PD}{PF} $,且\[ \frac{PD}{ED}\cdot \frac{EF}{PF}=\frac{\sin \angle PED}{\sin \angle DPE}\cdot \frac{\sin \angle FPE}{\sin \angle FEP}=\frac{\sin \angle BCP}{\sin \angle ACB}\cdot \frac{\sin \angle BAC}{\sin \angle BCP}=\frac{BC}{BA} \] |
|