Giải nhanh giúp mk

Bài 4 a) Ta có tứ giác ADME có ba góc vuông nên là hình vuông. b) Ta có: $\frac{AM}{AC}=\frac{AB}{BC}$ $\frac{AM}{AC}.\frac{AC}{AB}=\frac{AB}{BC}.\frac{AC}{AB}$ $\frac{AM}{AB}=\frac{AC}{BC}$ $\frac{AM}{AB}.\frac{AB}{AC}=\frac{AC}{BC}.\frac{AB}{AC}$ $\frac{AM}{AC}=\frac{AB}{BC}$ $AM.BC=AB.AC$ c) Ta có: $\widehat{AED}+\widehat{EMC}=90^{\circ}+\widehat{EMC}=180^{\circ}$ $\widehat{EMC}+\widehat{MCE}=90^{\circ}$ $\widehat{AED}+\widehat{MCE}=180^{\circ}$ $\widehat{DEM}+\widehat{MCE}=180^{\circ}$ $\widehat{DEM}+\widehat{MCE}+\widehat{DEC}=180^{\circ}$ $\widehat{DEC}=90^{\circ}$ $\Delta DEC$ vuông tại E Ta có: $\widehat{DEC}=\widehat{DIE}=90^{\circ}$ $\widehat{IDE}=\widehat{IDE}$ $\Delta DEC=\Delta DIE$ (g-g) $\widehat{DEI}=\widehat{DEC}=90^{\circ}$ $\Delta DEI$ vuông tại E d) Ta có: $\Delta DEC=\Delta DIE$ $DE=DI$ $DE=2EI$ $DI=2EI$ $\frac{DI}{EI}=2$ $\frac{CM}{MI}=2$ $\frac{CM}{CI}=\frac{2}{3}$ $\frac{AM}{CM}=\frac{2}{3}$ $\frac{AM}{AC}=\frac{2}{3}$ $\frac{AB}{BC}=\frac{2}{3}$ $BC=\frac{3}{2}.AB$ Vậy $\Delta ABC$ phải có thêm điều kiện $BC=\frac{3}{2}.AB$ thì $DE=2EI$.