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Question-207486




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

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