Giải thích các bước giải:

a.Xét $\Delta ABC,\Delta MCD$ có:

Chung $\hat C$

$\widehat{CAB}=\widehat{CDM}(=90^o)$ 

$\to\Delta ABC\sim\Delta DMC(g.g)$

Ta có $\Delta ABC$ vuông tại $A\to AC=\sqrt{BC^2-AB^2}=8$

Mà $\Delta ABC\sim\Delta DMC$

$\to \dfrac{DM}{AB}=\dfrac{MC}{BC}=\dfrac{CD}{AC}$

$\to \dfrac{DM}{6}=\dfrac{MC}{10}=\dfrac{3}8$

$\to DM=\dfrac94, MC=\dfrac{15}4$

$\to MB=BC-MC=\dfrac{25}4$

b.Xét $\Delta ABC,\Delta BME$ có:

Chung $\hat B$

$\widehat{BAC}=\widehat{BME}(=90^o)$

$\to\Delta BAC\sim\Delta BME(g.g)$

c.Từ câu b$\to \dfrac{BA}{BM}=\dfrac{BC}{BE}$

$\to \dfrac{BA}{BC}=\dfrac{BM}{BE}$

Mà $\widehat{ABM}=\widehat{EBC}$

$\to\Delta BAM\sim\Delta BCE(c.g.c)$

$\to \widehat{BMA}=\widehat{BEC}$

d.Từ câu c $\to \dfrac{BA}{BC}=\dfrac{BM}{BE}$

$\to BA.BE=BM.BC=\dfrac{125}{2}$

Ta có: $CD.CA=3.8=24$

$\to BA.BE+CD.CA=\dfrac{173}2\ne BC^2$