giải chi tiết ra

Bài 4. a) Ta có $\widehat{AHD}=\widehat{AKD}=90^{\circ}$ nên tứ giác AHKD nội tiếp. $\Rightarrow \widehat{HAD}=\widehat{HKD}$ (cung HD) $\widehat{HDA}=\widehat{KDE}$ (đối đỉnh) $\Rightarrow \Delta AHD \sim \Delta EKD$ (g-g) $\Rightarrow \frac{HD}{KD}=\frac{AH}{KE}=\frac{7,5}{6}=\frac{5}{4}$ $\Rightarrow HD=\frac{5}{4}\times KD$ Mặt khác ta có $\widehat{AED}=\widehat{KED}=90^{\circ},~\widehat{ADE}=\widehat{KDE}$ (đối đỉnh) $\Rightarrow \Delta AED \sim \Delta KED$ (g-g) $\Rightarrow \frac{AE}{KE}=\frac{DE}{DE}$ $\Rightarrow DE=\frac{5}{6}\times KE=5(cm)$ Ta có $AD=AE+DE=5+5=10(cm)$ $\Rightarrow HD=\frac{5}{4}\times 10=12,5(cm)$ b) Ta có $\widehat{AHD}=\widehat{AKD}=90^{\circ}$ nên tứ giác AHKD nội tiếp. $\Rightarrow \widehat{HAD}=\widehat{HKD}$ (cung HD) $\widehat{HDA}=\widehat{KDE}$ (đối đỉnh) $\Rightarrow \Delta AHD \sim \Delta EKD$ (g-g) $\Rightarrow \frac{HD}{KD}=\frac{AH}{KE}=\frac{7,5}{6}=\frac{5}{4}$ $\Rightarrow HD=\frac{5}{4}\times KD$ Mặt khác ta có $\widehat{AED}=\widehat{KED}=90^{\circ},~\widehat{ADE}=\widehat{KDE}$ (đối đỉnh) $\Rightarrow \Delta AED \sim \Delta KED$ (g-g) $\Rightarrow \frac{AE}{KE}=\frac{DE}{DE}$ $\Rightarrow DE=\frac{5}{6}\times KE=5(cm)$ Ta có $AD=AE+DE=5+5=10(cm)$ $\Rightarrow HD=\frac{5}{4}\times 10=12,5(cm)$ c) Ta có $\widehat{AHD}=\widehat{AKD}=90^{\circ}$ nên tứ giác AHKD nội tiếp. $\Rightarrow \widehat{HAD}=\widehat{HKD}$ (cung HD) $\widehat{HDA}=\widehat{KDE}$ (đối đỉnh) $\Rightarrow \Delta AHD \sim \Delta EKD$ (g-g) $\Rightarrow \frac{HD}{KD}=\frac{AH}{KE}=\frac{7,5}{6}=\frac{5}{4}$ $\Rightarrow HD=\frac{5}{4}\times KD$ Mặt khác ta có $\widehat{AED}=\widehat{KED}=90^{\circ},~\widehat{ADE}=\widehat{KDE}$ (đối đỉnh) $\Rightarrow \Delta AED \sim \Delta KED$ (g-g) $\Rightarrow \frac{AE}{KE}=\frac{DE}{DE}$ $\Rightarrow DE=\frac{5}{6}\times KE=5(cm)$ Ta có $AD=AE+DE=5+5=10(cm)$ $\Rightarrow HD=\frac{5}{4}\times 10=12,5(cm)$