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Let $O$ be any point inside $\Delta ABC$. Extend $AO,BO,CO$ to meet $BC,CA,AB$ at $D,E,F$ respectively.

By John Peck
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Let $O$ be any point inside $\Delta ABC$. Extend $AO,BO,CO$ to meet $BC,CA,AB$ at $D,E,F$ respectively. If $AO = 30, FO = 20, BO = 60, DO = 10, CO = 20$, find $EO$.

What I Tried: Here is a picture in Geogebra :-

The statement of the problem is really simple, every length surrounding point $O$ is given except $EO$ , which we have to find. My question is, how?

I am only given the lengths, which I don't even know how to use them. They neither help in finding lengths of other unknown lengths, nor help in angle-chasing, and so on. I don't find any similar triangles where I could have used the lengths. The only thing which I find interesting is $FO = CO$. Maybe we can use them in the areas of the triangles considering them as bases, but I can't see to find a way out there. Also that dosen't make the point $O$ special, does it, or am I missing something?

Can anyone help me or give me some hints? Thank You.

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2 Answers

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$$\frac{\triangle OBC} {\triangle ABC} =\frac{OD} {AD} , \frac{\triangle AOC} {\triangle ABC} = \frac{OE} {BE} , \frac{\triangle AOB} {\triangle ABC} =\frac{OF} {FC}$$$$\frac{\triangle OBC} {\triangle ABC} + \frac{\triangle AOC} {\triangle ABC} +\frac{\triangle AOB} {\triangle ABC} = \frac{OD} {AD} +\frac{OE} {BE} +\frac{OF} {FC} =1$$

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Hint: Apply Melanaus theorem in $\Delta ABD,\Delta CBF,\Delta ADC,\Delta AFC,\Delta CFA,\Delta BEA$

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