Question

Geometry question

Answer 1

Whats the question

Answer 2

\(\dfrac{BP}{PC}=\dfrac{2}{3}\\
\dfrac{AP}{PD}=\dfrac{3}{5}\\\)

Find ratio of area of \(\dfrac{\triangle ABP}{\triangle DPC}\)

Answer 3

Im sorry idk

Answer 4

@iGreen

Answer 5

\[\frac{A_{APB}}{A_{DPC}} = \frac{ \frac{1}{2} \left\| \vec{PB} \times \vec{PA} \right\|}{\frac{1}{2} \left\| \vec{PC} \times \vec{PD} \right\|} = \frac{ |\vec{PB}| | \vec{PA}| \sin \theta}{ |\vec{PC}| | \vec{PD} | \sin \theta} = \frac{ \frac{2}{5} |\vec{CB}| \frac{3}{8}| \vec{DA}| }{ \frac{3}{5} |\vec{BC}|\frac{5}{8} | \vec{DA} | } = \ \frac{6}{15} \ = \frac{2}{5}\]