1,

\(\Delta CDE\) và \(\Delta CAB\) có:

\(\widehat{C}\) chung, \(\widehat{CDE}=\widehat{CAB}=90^o\)

\(\to\Delta CDE\backsim\Delta CAB\) (g.g)

2,

\(\Delta CDE\backsim \Delta CAB\to \dfrac{CD}{CE}=\dfrac{CA}{CB}\to \dfrac{CD}{CA}=\dfrac{CE}{BC}\)

\(\delta BEC\) và \(\Delta ADC\) có:

\(\widehat{C}\) chung, \(\dfrac{CE}{BC}=\dfrac{CD}{CA}\)

\(\to\Delta BEC\backsim\Delta ADC\) (c.g.c)

\(\to \dfrac{BE}{BC}=\dfrac{AD}{AC}\to BE.AC=AD.BC\)

3,

\(\Delta BEC\backsim\Delta ADC\to \widehat{BEC}=\widehat{ADC}=135^o\\\to \widehat{AEB}=45^o\)

\(\to \Delta ABE\) vuông cân tại \(A\)

\(AB^2+AE^2=BE^2\\\to 2AB^2=BE^2\\\to BE=\sqrt{2}AB\)

\(BM=\dfrac{1}{2}BE\to \dfrac{BM}{BC}=\dfrac{BE}{2BC}=\dfrac{AD}{2AC}=\dfrac{AH}{\sqrt{2}AC}\)

\(\Delta AHC\) và \(\Delta BHA\) có:

\(\widehat{AHC}=\widehat{BHA}=90^o,\widehat{ACH}=\widehat{BAH}\) (Phụ \(\widehat{HAC}\))

\(\to\Delta AHC\backsim\Delta BHA\) (g.g)

\(\to \dfrac{AH}{AC}=\dfrac{BH}{AB}\to \dfrac{BM}{BC}=\dfrac{BH}{\sqrt{2}AB}=\dfrac{BH}{BE}\to \dfrac{BM}{BH}=\dfrac{BC}{BE}\)

\(\Delta BHM\) và \(\Delta BEC\) có:

\(\widehat{B}\) chung, \(\dfrac{BM}{BH}=\dfrac{BC}{BE}\)

\(\to \Delta BHM\backsim\Delta BEC\) (c.g.c)

4,

\(\Delta ABE\) vuông cân tai \(A\) có \(AM\) là đường trung tuyến

\(\to AM\) là đường phân giác hay\(AG\) là đường phân giác

\(\to \dfrac{GB}{GC}=\dfrac{AB}{AC}\)

\(\Delta CDE\backsim CAB\to \dfrac{DE}{CD}=\dfrac{AB}{AC}\\\to \dfrac{GB}{GC}=\dfrac{DE}{CD}\\DE//AH(\bot BC)\to \dfrac{DE}{AH}=\dfrac{CD}{CH}\to\dfrac{DE}{CD}=\dfrac{AH}{CH}\\\to \dfrac{GB}{GC}=\dfrac{AH}{CH}\\\to \dfrac{GB}{GC+GB}=\dfrac{AH}{AH+HC}\\\to \dfrac{GB}{BC}=\dfrac{DH}{AB+HC}\)