# Find constants such that three points are collinear

Let $ABC$ be a triangle and consider two points $D$ and $E$ such that $D$ lies on $AB$, $E$ lies on $AC$, $\frac{AD}{DB} = k$ and $\frac{AE}{EC} = p$.

Prove that the middles of $(AB)$, $(AC)$ and $(DE)$ are collinear if and only if $p = \frac{1}{k}$.

I need a solution involving Menelau's theorem. I tried to apply this theorem in triangle $ADE$ but got nothing.

Thank you in advance!

• Is $D$ on $AB$ and $E$ on $AC$? – gambler101 Sep 5 '16 at 16:07
• Yes, you are right – George R. Sep 5 '16 at 16:09
• Oh, come on! Couldn't you write this in your post? This makes the problem elementary! – Futurologist Sep 5 '16 at 16:18

## 1 Answer

Let $M$ be the midpoint of $AB$ and $N$ - the midpoint of $AC$. Then apply Menelaus' theorem to triangle $ADE$ and the line $MN$.

Let $MN$ intersect $DE$ in point $P$. Then by Menelaus' theorem $$\frac{AM}{MD} \cdot \frac{DP}{PE} \cdot \frac{EN}{NA} = 1$$ which can be written as $$\frac{PE}{DP} = \frac{AM}{MD}\cdot \frac{EN}{NA}$$ But $AM = MB = \frac{1}{2} AB$ and since $\frac{AD}{DB} = k$, $$\frac{AM}{MD} = \frac{\frac{1}{2}AB}{\frac{1}{2}AB - AD} = \frac{\frac{1}{2}(AD + DB)}{\frac{1}{2}(AD+DB) - AD} = \frac{AD + DB}{AD+DB - 2 \, AD} = \frac{AD + DB}{DB - AD} = \left( \frac{AD+ DB}{DB} \right) \,\left( \frac{1}{1-\frac{AD}{DB}}\right) = \left( \frac{AD}{DB} + 1 \right) \,\left( \frac{1}{1-\frac{AD}{DB}}\right) = \frac{1+k}{1-k}.$$ Furthermore, $AN = CN = \frac{1}{2} AC$ and since $\frac{AE}{EC} = p$, $$\frac{EN}{NA} = \frac{CN - EC}{NA} = \frac{\frac{1}{2}AC - EC}{NA} = \frac{\frac{1}{2}(AE+EC) - EC}{NA} = \frac{AE - EC}{2 \, NA} = \frac{AE - EC}{AC} = \frac{AE - EC}{AE+EC} = \frac{p-1}{p+1}.$$

Finally, $P$ is the midpoint of $DE$ if and only if $$1 = \frac{PE}{DP} = \frac{AM}{MD}\cdot \frac{EN}{NA} = \left( \frac{1+k}{1-k} \right) \cdot \left(\frac{p-1}{p+1}\right)$$ i.e. $$\frac{1-k}{1+k} = \frac{p-1}{p+1}$$ $$(1-k) (p+1) = (1+k) (p-1)$$ $$p+1 - kp - k = p-1 + kp - k$$ $$2 = 2 \, kp$$ $$p = \frac{1}{k}$$