Geometry Attack Attack

Question

mathmath333 · Answer

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$\large \color{black}{\begin{align} \dfrac{QD}{DR}=\dfrac{2}{3}\hspace{.33em}\~\
PC=CR\hspace{.33em}\~\
\end{align}}$

mathmath333 · Answer

Find

 $\large \color{black}{\begin{align} \dfrac{Area(\triangle AQD)}{Area(\triangle PAC)}=?\hspace{.33em}\~\

\end{align}}$

isaiah.feynman · Answer

Cevian Theorem?

isaiah.feynman · Answer

I don't recall it right now.

mathmath333 · Answer

but that theorem has like 3 medial like lines , the figure has 2, any ways thnx for sharing

rational · Answer

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isaiah.feynman · Answer

All I remember is that in a triangle,

rational · Answer

just an observation Area($\triangle PQC$) = Area($\triangle QCR$)

isaiah.feynman · Answer

The proof of the theorem makes use of the areas of the little triangles inside the big triangle.

rational · Answer

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rational · Answer

Few observations 
1) Area($\triangle PQC$) = Area($\triangle QCR$) 
2) Area($\triangle PDC$) = Area($\triangle DCR$) 
3) Area($\triangle QCD$)/Area($\triangle DCR$) = 2/3 
4) Area($\triangle QPD$)/Area($\triangle DPR$) = 2/3

isaiah.feynman · Answer

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isaiah.feynman · Answer

Then (RAC) = (CAP). (AQD)/(ADR) = 2/3. (AQD) = 2/3(ADR). Then, (CAP)/(AQD) = 3(PAC)/2(ADR)...

mathmath333 · Answer

answer is $\Large \dfrac{8}{15}$

isaiah.feynman · Answer

I don't think what I did is right.

isaiah.feynman · Answer

8 and 15 aren't multiples of 3 and 2. Perhaps it should be flipped.

rational · Answer

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