Giải thích các bước giải:

1.Xét $\Delta ABD,\Delta ACE$ có:

Chung $\hat A$

$\widehat{ADB}=\widehat{AEC}(=90^o)$

$\to \Delta ABD\sim\Delta ACE(g.g)$

$\to \dfrac{AB}{AC}=\dfrac{AD}{AE}$

$\to AB\cdot AE=AD\cdot AC$

2.Xét $\Delta ADE,\Delta ABC$ có:

Chung $\hat A$

$\dfrac{AD}{AB}=\dfrac{AE}{AC}$

$\to \Delta ADE\sim \Delta ABC(c.g.c)$

3.Xét $\Delta IBE,\Delta IDC$ có:

Chung $\hat I$

$\widehat{IEB}=\widehat{AED}=\widehat{ACB}=\widehat{ICD}$

$\to \Delta IBE\sim\Delta IDC(g.g)$

$\to \dfrac{IB}{ID}=\dfrac{IE}{IC}$

$\to IE\cdot ID=IB\cdot IC=(IM-BM)(IM+CM)=(IM-CM)(IM+CM)=IM^2-CM^2$

4.Gọi $AH\cap BC=F$

   Ta có: $BD\perp AC, CE\perp AB, BD\cap CE=H\to H$ là trực tâm $\Delta ABC$

$\to AH\perp BC=F$

Xét $\Delta BFH,\Delta BDC$ có:

Chung $\hat B$

$\widehat{BFH}=\widehat{BDC}(=90^o)$

$\to \Delta BFH\sim\Delta BDC(g.g)$

$\to \dfrac{BF}{BD}=\dfrac{BH}{BC}$

$\to BH\cdot BD=BF\cdot BC$

Tương tự: $CH\cdot CE=CF\cdot CB$

$\to P=BH\cdot BD+CH\cdot CE=BF\cdot BC+CF\cdot BC=BC(BF+FC)=BC^2$