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Question Number 207486 by hardmath last updated on 17/May/24

Commented by som(math1967) last updated on 17/May/24

$$DE\mathrm{\perp }AK?$$

Commented by hardmath last updated on 17/May/24

$$\mathrm{yes}\mathrm{professor}$$

Answered by A5T last updated on 17/May/24

$$\frac{AB}{BD}=\frac{AK}{KD}\Rightarrow \frac{BD}{KD}=\frac{7}{8}$$$$\frac{DE}{BC}=\frac{KD}{KB}=\frac{8}{15}\Rightarrow DE=16$$

Answered by som(math1967) last updated on 17/May/24

$$\frac{AB}{AK}=\frac{BD}{DK}=\frac{7}{8}\left[ADisbisectorof\mathrm{\angle }A\right]$$$$\Rightarrow \frac{BD}{DK}=\frac{7}{8}$$$$\Rightarrow \frac{DK}{BK}=\frac{8}{8+7}=\frac{8}{15}$$$$△KDE\sim △KBC$$$$\therefore \frac{DE}{BC}=\frac{DK}{BK}$$$$\Rightarrow \frac{x}{30}=\frac{8}{15}\therefore x=16$$

Commented by hardmath last updated on 17/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{professor}$$

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