Giải thích các bước giải:
a.Xét $\Delta ABC,\Delta MDC$ có:
Chung $\hat C$
$\widehat{BAC}=\widehat{DMC}=90^o$
$\to\Delta ABC\sim\Delta MDC(g.g)$
b.Ta có $\Delta ABC$ vuông tại $A$
$\to BC=\sqrt{AB^2+AC^2}=30\to MB=MC=\dfrac12BC=15$ vì $M$ là trung điểm $BC$
Mà $\Delta ABC\sim\Delta MDC$
$\to \dfrac{DC}{BC}=\dfrac{MD}{AB}=\dfrac{MC}{AC}$
$\to \dfrac{DC}{30}=\dfrac{MD}{18}=\dfrac{15}{24}$
$\to CD=\dfrac{75}{4}, MD=\dfrac{45}{4}$
c.Gọi $BD\cap EC=F$
Ta có $EM\perp BC, CA\perp BE\to D$ là trực tâm $\Delta EBC$
$\to BD\perp EC\to BF\perp EC$
Xét $\Delta EAC,\Delta EFB$ có:
Chung $\hat E$
$\widehat{EAC}=\widehat{EFB}=90^o$
$\to\Delta EAC\sim\Delta EFB(g.g)$
$\to\dfrac{EA}{EF}=\dfrac{EC}{EB}$
$\to EA.EB=EF.EC$
Xét $\Delta CME,\Delta CFB$ có:
Chung $\hat C$
$\widehat{CME}=\widehat{CFB}=90^o$
$\to\Delta CME\sim\Delta CFB(g.g)$
$\to\dfrac{CM}{CF}=\dfrac{CE}{CB}$
$\to CM.CB=CF.CE$
$\to CM.CB+EA.EB=EF.EC+CF.CE=CE^2$
