Giải thích các bước giải:

a.Xét $\Delta EBD,\Delta EAC$ có:

Chung $\hat E$

$\widehat{EDB}=\widehat{EAC}(=90^o)$

$\to \Delta EBD\sim\Delta ECA(g.g)$

$\to \dfrac{EB}{EC}=\dfrac{ED}{EA}$

$\to EA\cdot EB=ED\cdot EC$

$\to \dfrac{EA}{ED}=\dfrac{EC}{EA}$

Mà $\widehat{AED}=\widehat{BEC}$

$\to \Delta EAD\sim\Delta ECB(c.g.c)$

$\to \widehat{EAD}=\widehat{ECB}$

b.Ta có:

$\widehat{AMD}=\widehat{BMC}=120^o$

$\to \widehat{AED}=360^o-\widehat{EAM}-\widehat{EDM}-\widehat{AMD}=60^o$

$\to \widehat{BED}=60^o\to \Delta BED$ là nửa tam giác đều cạnh $BE$

$\to \dfrac{DE}{BE}=\dfrac12$

Từ câu a $\to \dfrac{S_{EAD}}{S_{EBC}}=(\dfrac{ED}{EB})^2=\dfrac14$

$\to S_{EBC}=4S_{AED}=144$

c.Gọi $EM\cap BC=F$

Ta có: $CA\perp BE, BD\perp CE, BD\cap CA=M\to M$ là trực tâm $\Delta EBC\to EF\perp CB$

Xét $\Delta BMF,\Delta BDC$ có:

Chung $\hat B$

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

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

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

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

Tương tự $CM\cdot CA=CF\cdot CB$

$\to BM\cdot BD+CM\cdot CA=BF\cdot BC=CF\cdot BC=BC^2$

d.Ta có: $P, Q$ là trung điểm $HB, HD\to PQ$ là đường trung bình $\Delta HBD$

$\to PQ//DB$

Mà $BD\perp DC\to PQ\perp CD$

       $DH\perp BC\to DQ\perp CP$

$\to Q$ là trực tâm $\Delta PCD\to CQ\perp PD$