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+0  
 
+5
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avatar+1718 

 

I treid drawing a picture then promptly got stuck... Halp!

 

- A very confused

\(tommarvoloriddle\)

 Jul 5, 2019
 #1
avatar+7725 
+3

Connect AK and extend it to meet BC at F.

By Ceva's theorem,

\(\dfrac{\text{AE}}{\text{EB}} \cdot \dfrac{\text{BF}}{\text{FC}} \cdot \dfrac{\text{CD}}{\text{DA}} = 1\\ \text{Let }\dfrac{\text{BF}}{\text{FC}} = k,\\ \dfrac{2}{3} \cdot k \cdot \dfrac{1}{2} = 1\\ k = 3\\ \text{BF}:\text{FC} = 3:1\)

By Menelaus' theorem,

\(\dfrac{\text{BF}}{\text{FC}} \cdot \dfrac{\text{CA}}{\text{AD}}\cdot \dfrac{\text{DK}}{\text{KB}} = 1\\ 3\cdot \dfrac{3}{2} \cdot \dfrac{\text{DK}}{\text{KB}} = 1\\ \boxed{\dfrac{\text{DK}}{\text{KB}} = \color{red}{\dfrac{2}{9}}}\)

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 Jul 5, 2019
 #2
avatar+1718 
+8

Yay! Thank you, MaxWong!

tommarvoloriddle  Jul 5, 2019

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