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# geometry

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In triangle ABC, points D and E lie on BC and AC, respectively. If AD and BE intersect at T so that AT/DT = 3 and BT/ET = 4, what is CD/BD?

Jun 24, 2021

#1
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CD / BD = 4/11

Jun 24, 2021
#2
+26117
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In triangle ABC, points D and E lie on BC and AC, respectively.

If AD and BE intersect at T so that AT/DT = 3 and BT/ET = 4, what is CD/BD?

$$\text{Let \vec{CA}=\vec{b}} \\ \text{Let \vec{CB}=\vec{a}} \\ \text{Let \mu = \dfrac{3}{4} = \dfrac{AT}{AD}} \\ \text{Let 1-\mu = \dfrac{1}{4} = \dfrac{DT}{AD}} \\ \text{Let \rho = \dfrac{4}{5} = \dfrac{BT}{BE}} \\ \text{Let 1-\rho = \dfrac{1}{5} = \dfrac{ET}{BE}} \\ \text{Let \lambda = \dfrac{BD}{BC} } \\ \text{Let 1-\lambda = \dfrac{CD}{BC} } \\ \text{Let \mathbf{ \dfrac{1-\lambda}{\lambda} = \dfrac{CD}{BD}} } \\ \text{Let \epsilon = \dfrac{CE}{CA} }$$

$$\text{In C-D-T-E} \\ \begin{array}{|rcll|} \hline (1-\lambda)\vec{a}+(1-\mu)[\vec{b}-(1-\lambda)\vec{a}] &=& \epsilon \vec{b} + (1-\rho)(\vec{a}-\epsilon \vec{b}) \\ (1-\lambda)\vec{a}+(1-\mu)\vec{b} -(1-\mu)(1-\lambda)\vec{a} &=& \epsilon \vec{b} + (1-\rho)\vec{a} - (1-\rho)\epsilon \vec{b}) \\ \vec{a}[(1-\lambda)-(1-\mu)(1-\lambda)-(1-\rho) ] &=& \vec{b} [\epsilon - (1-\rho)\epsilon -(1-\mu) ] \\ \vec{a}[(1-\lambda)\Big(1-(1-\mu)\Big)-(1-\rho) ] &=& \vec{b} [\epsilon - (1-\rho)\epsilon -(1-\mu) ] \\ \vec{a}[\underbrace{ (1-\lambda)\mu)-(1-\rho) }_{=0} ] &=& \vec{b} [\underbrace{ \epsilon - (1-\rho)\epsilon -(1-\mu) }_{=0} ] \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline (1-\lambda)\mu-(1-\rho) &=& 0 \\ (1-\lambda)\mu &=& (1-\rho) \\\\ \mathbf{ 1-\lambda } &=& \mathbf{\dfrac{(1-\rho)}{\mu} } \\ \hline 1-\lambda &=& \dfrac{(1-\rho)}{\mu} \\ \lambda &=& 1-\dfrac{(1-\rho)}{\mu} \\ \mathbf{ \lambda} &=& \mathbf{ \dfrac{\mu-(1-\rho)}{\mu} } \\ \hline \dfrac{CD}{BD} &=& \dfrac{1-\lambda}{\lambda} \\ \dfrac{CD}{BD} &=& \dfrac{\dfrac{(1-\rho)}{\mu}} {\dfrac{\mu-(1-\rho)}{\mu}} \\ \\ \dfrac{CD}{BD} &=& \dfrac{(1-\rho)} {\mu-(1-\rho)} \\ \\ \dfrac{CD}{BD} &=& \dfrac{\dfrac{1}{5}} {\dfrac{3}{4}-\dfrac{1}{5}} \\ \\ \dfrac{CD}{BD} &=& \dfrac{\dfrac{1}{5}} {\dfrac{11}{20}} \\ \\ \dfrac{CD}{BD} &=& \dfrac{1}{5} * \dfrac{20}{11} \\ \\ \mathbf{ \dfrac{CD}{BD}} &=& \mathbf{\dfrac{4}{11}} \\ \hline \end{array}$$

Jun 24, 2021