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I have the following question with me:

In a triangle ABC D and E lie on sides BC and AB respectively. F lies on AC such that EF is parallel to BC, G lies on side BC such that EG lies parallel to AD. Let M and N be midpoints of AD and BC respectively. Prove that the midpoint of GF lies on line MN. Also prove that $\frac{EF}{BC} + \frac{EG}{AD} = 1$

How do I solve this question? I do not want to use coordinate geometry or vectors to solve this euqstion.

The only thing I figured out is that FG is also parallel to AB. Does it help?

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Let $\frac{AF}{AC}=\frac{AE}{AB}=x$ and $K$ be a midpoint of $FG$.

Thus, $$\frac{BG}{BD}=\frac{BE}{AB}=1-x$$ and from here $$\vec{KN}=\frac{1}{2}\left(\vec{FC}+\vec{GB}\right)=\frac{1}{2}\left((1-x)\vec{AB}+(1-x)\vec{DB}\right)=\frac{1}{2}(1-x)\vec{MN},$$ which says $K\in MN.$

The second is true because $$\frac{EF}{BC}+\frac{EG}{AD}=\frac{AE}{AB}+\frac{BE}{AB}=1.$$

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  • $\begingroup$ thanks for the help, vectors did simplify a lot of stuff. However, as I mentioned can I get a solution without coordinate geometry or vectors? $\endgroup$ – saisanjeev Jan 6 at 9:53

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