# How can I find ratio between area of triangle and area of quadrilateral?

I have a parallelogram $ABCD$. $E$ is center of $AD$. $O$ is center of $AC$ and center of $DB$. $F$ is the intersection point between $CE$ and $DO$. Point $G$ is the intersection between $EO$ and $AF$. $H$ is the middle of $DC$.

I put variables to different lines in the figure to show the ratios I found.

I need to find the ratio

$$Area_{AEG} : Area_{AEFO}$$

I don't have an idea on how to find it. I know how to find ratios like for example $$Area_{AEG} : Area_{ADH}$$ because they are similar triangles.

If $H$ is the intersection between $AF$ and $CD$, then $GF=(1/2)FH$ so that $GF=(1/3)GH$, that is: $GF=(1/3)AG$. If $AM$ and $FN$ are the altitudes of triangles $AEO$ and $FEO$ with respect to $EO$ as common base, then $AGM$ and $FGN$ are similar triangle and $FN=(1/3)AM$. It follows that $Area_{EOF}={1\over3}Area_{EOA}$.

On the other hand triangles $AEG$ and $AOG$ have equal bases $EG$ and $OG$, and share the same altitude $AM$. It follows that $Area_{AEG}=Area_{AOG}$, that is: $Area_{AEG}={1\over2}Area_{EOA}$.

In the end we have: $$Area_{AEFO}=Area_{EOF}+Area_{EOA}={1\over3}Area_{EOA}+Area_{EOA}={4\over3}Area_{EOA}$$ and $$Area_{AEG}={1\over2}Area_{EOA},$$ so that $${Area_{AEG}\over Area_{AEFO}}= {(1/2)Area_{EOA}\over(4/3)Area_{EOA}}={3\over8}.$$

• Your answer is helping me to resolve it, however, the ratio between similar triangles is not the same than the ratio between their areas. – Pichi Wuana May 10 '16 at 19:36
• You know that there need to be equal angles between similar triangles... – Pichi Wuana May 10 '16 at 19:45
• I'm not using similar triangles. If triangles $EOF$ and $EOA$ share the same basis and their altitudes are in the ratio $1:3$, then their areas are in the same ratio. – Aretino May 10 '16 at 19:51
• OK, I've expanded my answer. – Aretino May 10 '16 at 21:27
• You are right: I inadvertedly wrote the inverse fraction. Corrected now. – Aretino May 10 '16 at 22:19