Question Number 207486 by hardmath last updated on 17/May/24
Commented by som(math1967) last updated on 17/May/24
$${DE}\bot{AK}\:? \\ $$
Commented by hardmath last updated on 17/May/24
$$\mathrm{yes}\:\mathrm{professor} \\ $$
Answered by A5T last updated on 17/May/24
$$\frac{{AB}}{{BD}}=\frac{{AK}}{{KD}}\Rightarrow\frac{{BD}}{{KD}}=\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\frac{{DE}}{{BC}}=\frac{{KD}}{{KB}}=\frac{\mathrm{8}}{\mathrm{15}}\Rightarrow{DE}=\mathrm{16} \\ $$
Answered by som(math1967) last updated on 17/May/24
![((AB)/(AK))=((BD)/(DK)) =(7/8) [AD is bisector of∠A] ⇒((BD)/(DK))=(7/8) ⇒((DK)/(BK))=(8/(8+7))=(8/(15)) △KDE∼△KBC ∴ ((DE)/(BC))=((DK)/(BK)) ⇒(x/(30))=(8/(15)) ∴x=16](https://www.tinkutara.com/question/Q207494.png)
$$\:\frac{{AB}}{{AK}}=\frac{{BD}}{{DK}}\:=\frac{\mathrm{7}}{\mathrm{8}}\:\:\:\:\left[{AD}\:{is}\:{bisector}\:{of}\angle{A}\right] \\ $$$$\:\Rightarrow\frac{{BD}}{{DK}}=\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\Rightarrow\frac{{DK}}{{BK}}=\frac{\mathrm{8}}{\mathrm{8}+\mathrm{7}}=\frac{\mathrm{8}}{\mathrm{15}} \\ $$$$\:\bigtriangleup{KDE}\sim\bigtriangleup{KBC} \\ $$$$\therefore\:\frac{{DE}}{{BC}}=\frac{{DK}}{{BK}} \\ $$$$\Rightarrow\frac{{x}}{\mathrm{30}}=\frac{\mathrm{8}}{\mathrm{15}}\:\:\:\therefore{x}=\mathrm{16} \\ $$
Commented by hardmath last updated on 17/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor} \\ $$